Trong bài này, chúng ta tìm hiểu một số bài toán tìm nguyên hàm của hàm lượng giác có dạng khá đặc biệt.
I. Dạng 1. $I = \int {\frac{{dx}}{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}} $
1. Phương pháp tính
Dùng đồng nhất thức:
$1 = \frac{{\sin \left( {a - b} \right)}}{{\sin \left( {a - b} \right)}} = \frac{{\sin \left[ {\left( {x + a} \right) - \left( {x + b} \right)} \right]}}{{\sin \left( {a - b} \right)}} = \frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\sin \left( {a - b} \right)}}$
Từ đó suy ra:
$I = \frac{1}{{\sin \left( {a - b} \right)}}\int {\frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}} dx$
$ = \frac{1}{{\sin \left( {a - b} \right)}}\int {\left[ {\frac{{\cos \left( {x + b} \right)}}{{\sin \left( {x + b} \right)}} - \frac{{\cos \left( {x + a} \right)}}{{\sin \left( {x + a} \right)}}} \right]} dx$
$ = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\ln \left| {\sin \left( {x + b} \right)} \right| - \ln \left| {\sin \left( {x + a} \right)} \right|} \right] + C$
2. Chú ý
Với cách này, ta có thể tìm được các nguyên hàm:
•$J = \int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} $ bằng cách dùng đồng nhất thức $1 = \frac{{\sin \left( {a - b} \right)}}{{\sin \left( {a - b} \right)}}$
•$K = \int {\frac{{dx}}{{\sin \left( {x + a} \right)\cos \left( {x + b} \right)}}} $ bằng cách dùng đồng nhất thức $1 = \frac{{\cos \left( {a - b} \right)}}{{\cos \left( {a - b} \right)}}$
3. Ví dụ áp dụng
• $I = \int {\frac{{dx}}{{\sin x\sin \left( {x + \frac{\pi }{6}} \right)}}} $
Ta có: $1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {x + \frac{\pi }{6}} \right) - x} \right]}}{{\frac{1}{2}}} = 2\left[ {\sin \left( {x + \frac{\pi }{6}} \right)\cos x - \cos \left( {x + \frac{\pi }{6}} \right)\sin x} \right]$
Từ đó: $I = 2\int {\frac{{\left[ {\sin \left( {x + \frac{\pi }{6}} \right)\cos x - \cos \left( {x + \frac{\pi }{6}} \right)\sin x} \right]}}{{\sin x\sin \left( {x + \frac{\pi }{6}} \right)}}} dx = 2\int {\left[ {\frac{{\cos x}}{{\sin x}} - \frac{{\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \right]} dx$
$ = 2\int {\frac{{d\left( {\sin x} \right)}}{{\sin x}}} - 2\int {\frac{{d\left( {\sin \left( {x + \frac{\pi }{6}} \right)} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = 2\ln \left| {\frac{{\sin x}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \right| + C$
$1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {3x + \frac{\pi }{6}} \right) - 3x} \right]}}{{\frac{1}{2}}} = 2\left[ {\sin \left( {3x + \frac{\pi }{6}} \right)\cos 3x - \cos \left( {3x + \frac{\pi }{6}} \right)\sin 3x} \right]$
Từ đó:
$I = 2\int {\frac{{\left[ {\sin \left( {3x + \frac{\pi }{6}} \right)\cos 3x - \cos \left( {3x + \frac{\pi }{6}} \right)\sin 3x} \right]}}{{\cos 3x\cos \left( {3x + \frac{\pi }{6}} \right)}}} dx = 2\int {\frac{{\sin \left( {3x + \frac{\pi }{6}} \right)}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}dx} - 2\int {\frac{{\sin 3x}}{{\cos 3x}}dx} $
$ = - \frac{2}{3}\int {\frac{{d\left( {\cos \left( {3x + \frac{\pi }{6}} \right)} \right)}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}} + \frac{2}{3}\int {\frac{{d\left( {\cos 3x} \right)}}{{\cos 3x}}} = \frac{2}{3}\ln \left| {\frac{{\cos 3x}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}} \right| + C$
$ = \sqrt 2 \left[ {\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{{12}}} \right)} \right]$
Từ đó: $I = \sqrt 2 \int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{{12}}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right)}}} dx$
$ = \sqrt 2 \int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}dx} + \sqrt 2 \int {\frac{{\sin \left( {x + \frac{\pi }{{12}}} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} dx$
$ = \sqrt 2 \int {\frac{{d\left( {\sin \left( {x + \frac{\pi }{3}} \right)} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} - \sqrt 2 \int {\frac{{d\left( {\cos \left( {x + \frac{\pi }{{12}}} \right)} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} = \sqrt 2 \ln \left| {\frac{{\sin \left( {x + \frac{\pi }{3}} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} \right| + C$
II. Dạng 2. $I = \int {\tan \left( {x + a} \right)\tan \left( {x + b} \right)dx} $
1. Phương pháp tính
Ta có: $\tan \left( {x + a} \right)\tan \left( {x + b} \right) = \frac{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}$
$ = \frac{{\sin \left( {x + a} \right)\sin \left( {x + b} \right) + \cos \left( {x + a} \right)\cos \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}} - 1 = \frac{{\cos \left( {a - b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}} - 1$
Từ đó: $I = \cos \left( {a - b} \right)\int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} - 1$
Đến đây ta gặp bài toán tìm nguyên hàm ở Dạng 1.
2. Chú ý
Với cách này, ta có thể tính được các nguyên hàm:
•$J = \int {\cot \left( {x + a} \right)\cot \left( {x + b} \right)dx} $
•$K = \int {\tan \left( {x + a} \right)\tan \left( {x + b} \right)dx} $
3. Ví dụ áp dụng
•$I = \int {\cot \left( {x + \frac{\pi }{3}} \right)\cot \left( {x + \frac{\pi }{6}} \right)dx} $
Ta có:
Ta có: $\cot \left( {x + \frac{\pi }{3}} \right)\cot \left( {x + \frac{\pi }{6}} \right) = \frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}$
$ = \frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1$
$ = \frac{{\cos \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1 = \frac{{\sqrt 3 }}{2}.\frac{1}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1$
Từ đó: $I = \frac{{\sqrt 3 }}{2}\int {\frac{1}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}dx} - \int {dx} = \frac{{\sqrt 3 }}{2}{I_1} - x + C$
Tính ${I_1} = \int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} $
Ta có: $1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\frac{1}{2}}}$
$ = 2\left[ {\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)} \right]$
Từ đó: ${I_1} = 2\int {\frac{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} dx$
$ = 2\int {\frac{{\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}dx} - 2\int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}dx} = 2\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| + C$
Suy ra: $I = \frac{{\sqrt 3 }}{2}.2\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| - x + C = \sqrt 3 \ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| - x + C$
$ = \frac{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1$
$ = \frac{{\sin \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1 = \frac{1}{2}.\frac{1}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1$
Từ đó: $K = \frac{1}{2}\int {\frac{1}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}dx} + \int {dx} = \frac{1}{2}{K_1} + x + C$
Đến đây, bằng cách tính ở Dạng 1, ta tính được:
${K_1} = \int {\frac{{dx}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{2}{{\sqrt 3 }}\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)}}} \right| + C$
Suy ra: $K = \frac{{\sqrt 3 }}{3}\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)}}} \right| + x + C$
III. Dạng 3. $I = \int {\frac{{dx}}{{a\sin x + b\cos x}}} $
1. Phương pháp tính
Có: $a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\sin x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\cos x} \right)$
$ \Rightarrow a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \sin \left( {x + \alpha } \right)$
$ \Rightarrow I = \frac{1}{{\sqrt {{a^2} + {b^2}} }}\int {\frac{{dx}}{{\sin \left( {x + \alpha } \right)}}} = \frac{1}{{\sqrt {{a^2} + {b^2}} }}\ln \left| {\tan \frac{{x + \alpha }}{2}} \right| + C$
2. Ví dụ áp dụng
$ = - \frac{1}{4}\ln \left| {\tan \frac{{\frac{\pi }{6} - 2x}}{2}} \right| + C = - \frac{1}{4}\ln \left| {\tan \left( {\frac{\pi }{{12}} - x} \right)} \right| + C$
IV. Dạng 4. $I = \int {\frac{{dx}}{{a\sin x + b\cos x + c}}} $
1. Phương pháp tính
Đặt $\tan \frac{x}{2} = t \Rightarrow \left\{ \begin{array}{l}
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}\\
\tan x = \frac{{2t}}{{1 - {t^2}}}
\end{array} \right.$
2. Ví dụ áp dụng
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
Từ đó: $I = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{3.\frac{{1 - {t^2}}}{{1 + {t^2}}} + 5\frac{{2t}}{{1 + {t^2}}} + 3}}} = \int {\frac{{2dt}}{{3 - 3{t^2} + 10t + 3 + 3{t^2}}}} = \int {\frac{{2dt}}{{10t + 6}}} $
$ = \frac{1}{5}\int {\frac{{d\left( {5t + 3} \right)}}{{5t + 3}}} = \frac{1}{5}\ln \left| {5t + 3} \right| + C = \frac{1}{5}\ln \left| {5\tan \frac{x}{2} + 3} \right| + C$
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
Từ đó: $J = \int {\frac{{2.\frac{{2dt}}{{1 + {t^2}}}}}{{2.\frac{{2t}}{{1 + {t^2}}} - \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{4dt}}{{4t - 1 + {t^2} + 1 + {t^2}}}} = \int {\frac{{4dt}}{{2{t^2} + 4t}}} = 2\int {\frac{{dt}}{{t\left( {t + 2} \right)}}} $
$ = \int {\left( {\frac{1}{t} - \frac{1}{{t + 2}}} \right)dt} = \ln \left| t \right| - \ln \left| {t + 2} \right| + C = \ln \left| {\tan \frac{x}{2}} \right| - \ln \left| {\tan \frac{x}{2} + 2} \right| + C$
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\tan x = \frac{{2t}}{{1 - {t^2}}}
\end{array} \right.$
Từ đó: $K = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{2t}}{{1 - {t^2}}}}}} = \frac{1}{2}\int {\frac{{1 - {t^2}}}{t}} dt = \frac{1}{2}\int {\frac{{dt}}{t}} - \frac{1}{2}\int {tdt} $
$ = \frac{1}{2}\ln \left| t \right| - \frac{1}{4}{t^2} + C = \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| - \frac{1}{4}{\tan ^2}\frac{x}{2} + C$
V. Dạng 5. $I = \int {\frac{{dx}}{{a.{{\sin }^2}x + b.\sin x\cos x + c.{{\cos }^2}x}}} $
1. Phương pháp tính
$I = \int {\frac{{dx}}{{\left( {a{{\tan }^2}x + b\tan x + c} \right).{{\cos }^2}x}}} $
Đặt $\tan x = t \Rightarrow \frac{{dx}}{{{{\cos }^2}x}} = dt$. Suy ra $I = \int {\frac{{dt}}{{a{t^2} + bt + c}}} $
2. Ví dụ áp dụng
$ \Rightarrow I = \int {\frac{{dt}}{{3{t^2} - 2t - 1}}} = \int {\frac{{dt}}{{\left( {t - 1} \right)\left( {3t + 1} \right)}}} $
$ = \frac{1}{4}\int {\left( {\frac{1}{{t - 1}} - \frac{3}{{3t + 1}}} \right)dt} = \frac{1}{4}\int {\frac{{dt}}{{t - 1}}} - \frac{1}{4}\int {\frac{{d\left( {3t + 1} \right)}}{{3t + 1}}} $
$ = \frac{1}{4}\ln \left| {\frac{{t - 1}}{{3t + 1}}} \right| + C = \frac{1}{4}\ln \left| {\frac{{\tan x - 1}}{{3\tan x + 1}}} \right| + C$
$ \Rightarrow J = \int {\frac{{dt}}{{{t^2} - 2t - 2}}} = \int {\frac{{d\left( {t - 1} \right)}}{{{{\left( {t - 1} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}} = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{t - 1 - \sqrt 3 }}{{t - 1 + \sqrt 3 }}} \right| + C$
$ = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{\tan x - 1 - \sqrt 3 }}{{\tan x - 1 + \sqrt 3 }}} \right| + C$
VI. Dạng 6. $I = \int {\frac{{{a_1}\sin x + {b_1}\cos x}}{{{a_2}\sin x + {b_2}\cos x}}dx} $
1. Phương pháp tính
Ta tìm A, B sao cho:
${a_1}\sin x + {b_1}\cos x = A\left( {{a_2}\sin x + {b_2}\cos x} \right) + B\left( {{a_2}\cos x - {b_2}\sin x} \right)$
2. Ví dụ áp dụng
$4\sin x + 3\cos x = A\left( {\sin x + 2\cos x} \right) + B\left( {\cos x - 2\sin x} \right)$
$ \Rightarrow 4\sin x + 3\cos x = \left( {A - 2B} \right)\sin x + \left( {2A + B} \right)\cos x \Rightarrow \left\{ \begin{array}{l}
A - 2B = 4\\
2A + B = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 2\\
B = - 1
\end{array} \right.$
Từ đó: $I = \int {\frac{{2\left( {\sin x + 2\cos x} \right) - \left( {\cos x - 2\sin x} \right)}}{{\sin x + 2\cos x}}} dx$
$ = 2\int {dx} - \int {\frac{{d\left( {\sin x + 2\cos x} \right)}}{{\sin x + 2\cos x}}} = 2x - \ln \left| {\sin x + 2\cos x} \right| + C$
$3\cos x - 2\sin x = A\left( {\cos x - 4\sin x} \right) + B\left( { - \sin x - 4\cos x} \right)$
$ \Rightarrow 3\cos x - 2\sin x = \left( {A - 4B} \right)\cos x + \left( { - 4A - B} \right)\sin x$
$ \Rightarrow \left\{ \begin{array}{l}
A - 4B = 3\\
4A + B = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = \frac{{11}}{{17}}\\
B = - \frac{{10}}{{17}}
\end{array} \right.$
Từ đó: $J = \int {\frac{{\frac{{11}}{{17}}\left( {\cos x - 4\sin x} \right) - \frac{{10}}{{17}}\left( { - \sin x - 4\cos x} \right)}}{{\cos x - 4\sin x}}} dx$
$ = \frac{{11}}{{17}}\int {dx} - \frac{{10}}{{17}}\int {\frac{{d\left( {\cos x - 4\sin x} \right)}}{{\cos x - 4\sin x}}} = \frac{{11}}{{17}}x - \frac{{10}}{{17}}\ln \left| {\cos x - 4\sin x} \right| + C$
3. Chú ý
Chẳng hạn:
$8\cos x = A\left( {\sqrt 3 \sin x + \cos x} \right) + B\left( {\sqrt 3 \cos x - \sin x} \right)$
$ \Rightarrow 8\cos x = \left( {A\sqrt 3 - B} \right)\sin x + \left( {A + B\sqrt 3 } \right)\cos x$
$ \Rightarrow \left\{ \begin{array}{l}
A\sqrt 3 - B = 0\\
A + B\sqrt 3 = 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 2\\
B = 2\sqrt 3
\end{array} \right.$
Từ đó: $I = \int {\frac{{2\left( {\sqrt 3 \sin x + \cos x} \right) + 2\sqrt 3 \left( {\sqrt 3 \cos x - \sin x} \right)}}{{{{\left( {\sqrt 3 \sin x + \cos x} \right)}^2}}}} dx$
$ = 2\int {\frac{{dx}}{{\sqrt 3 \sin x + \cos x}}} + 2\sqrt 3 \int {\frac{{d\left( {\sqrt 3 \sin x + \cos x} \right)}}{{{{\left( {\sqrt 3 \sin x + \cos x} \right)}^2}}}} = 2{I_1} - \frac{{2\sqrt 3 }}{{\sqrt 3 \sin x + \cos x}} + C$
Tìm ${I_1} = \int {\frac{{dx}}{{\sqrt 3 \sin x + \cos x}}} = \frac{1}{2}\int {\frac{{dx}}{{\frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x}}} = \frac{1}{2}\int {\frac{{dx}}{{\sin x\cos \frac{\pi }{6} + \cos x\sin \frac{\pi }{6}}}} $
$ = \frac{1}{2}\int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{1}{2}\int {\frac{{d\left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{1}{2}\ln \left| {\tan \frac{{x + \frac{\pi }{6}}}{2}} \right| + C = \frac{1}{2}\ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right)} \right| + C$
Vậy $I = \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right)} \right| - \frac{{2\sqrt 3 }}{{\sqrt 3 \sin x + \cos x}} + C$
$8\sin x + \cos x + 5 = A\left( {2\sin x - \cos x + 1} \right) + B\left( {2\cos x + \sin x} \right) + C$
$ \Rightarrow 8\sin x + \cos x + 5 = \left( {2A + B} \right)\sin x + \left( { - A + 2B} \right)\cos x + A + C$
$ \Rightarrow \left\{ \begin{array}{l}
2A + B = 8\\
- A + 2B = 1\\
A + C = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 3\\
B = 2\\
C = 2
\end{array} \right.$
Từ đó: $J = \int {\frac{{3\left( {2\sin x - \cos x + 1} \right) + 2\left( {2\cos x + \sin x} \right) + 2}}{{2\sin x - \cos x + 1}}} dx$
$ = 3\int {dx} + 2\int {\frac{{2\cos x + \sin x}}{{2\sin x - \cos x + 1}}} dx + 2\frac{{dx}}{{2\sin x - \cos x + 1}}$
$ = 3x + 2\ln \left| {2\sin x - \cos x + 1} \right| + 2{J_1}$
Tìm ${J_1} = \int {\frac{{dx}}{{2\sin x - \cos x + 1}}} $
Đặt $\tan \frac{x}{2} = t \Rightarrow \left\{ \begin{array}{l}
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
$ \Rightarrow {J_1} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{2.\frac{{2t}}{{1 + {t^2}}} - \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{dt}}{{{t^2} + 2t}}} = \int {\frac{{dt}}{{t\left( {t + 2} \right)}}} = \frac{1}{2}\int {\left( {\frac{1}{t} - \frac{1}{{t + 2}}} \right)} dt$
$ = \frac{1}{2}\ln \left| {\frac{t}{{t + 2}}} \right| + C = \frac{1}{2}\ln \left| {\frac{{\tan \frac{x}{2}}}{{\tan \frac{x}{2} + 2}}} \right| + C$
Vậy: $J = 3x + 2\ln \left| {2\sin x - \cos x + 1} \right| + \ln \left| {\frac{{\tan \frac{x}{2}}}{{\tan \frac{x}{2} + 2}}} \right| + C$
VII. Dạng 7. Biến đổi đưa về nguyên hàm cơ bản hoặc 6 dạng ở trên
$ = \frac{1}{4}\int {\sin 2xd\left( {\sin 2x} \right)} + \frac{1}{4}\int {\left( { - \sin 2x + \sin 6x} \right)} dx$
$ = \frac{1}{8}{\sin ^2}2x + \frac{1}{8}\cos 2x - \frac{1}{{24}}\cos 6x + C$
$ = \frac{{\sin x\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right)}}{{\cos x\left( {\cos 2x + \cos \frac{{2\pi }}{3}} \right)}} = \frac{{\sin x\left( {1 - 2{{\sin }^2}x + \frac{1}{2}} \right)}}{{\cos x\left( {2{{\cos }^2}x - 1 - \frac{1}{2}} \right)}}$
$ = \frac{{\sin x\left( {3 - 4{{\sin }^2}x} \right)}}{{\cos x\left( {4{{\cos }^2}x - 3} \right)}} = \frac{{3\sin x - 4{{\sin }^3}x}}{{4{{\cos }^3}x - 3\cos x}} = \frac{{\sin 3x}}{{\cos 3x}}$
Từ đó: $I = \int {\frac{{\sin 3x}}{{\cos 3x}}dx} = - \frac{1}{3}\int {\frac{{d\left( {\cos 3x} \right)}}{{\cos 3x}}} = - \frac{1}{3}\ln \left| {\cos 3x} \right| + C$
$ \Rightarrow {\sin ^3}x\sin 3x = \frac{{3\sin x - 4\sin 3x}}{4}.\sin 3x$
$ = \frac{3}{4}\sin x\sin 3x - \frac{1}{4}{\sin ^2}3x = \frac{3}{8}\left( {\cos 2x - \cos 4x} \right) - \frac{1}{8}\left( {1 - \cos 6x} \right)$
$ = \frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x - \frac{1}{8}$
Từ đó: $I = \int {\left( {\frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x - \frac{1}{8}} \right)} dx$
$ = \frac{3}{{16}}\sin 2x - \frac{3}{{32}}\sin 4x + \frac{1}{{48}}\sin 6x - \frac{1}{8}x + C$
${\cos ^3}x = \frac{{3\cos x + \cos 3x}}{4}$
Suy ra: ${\sin ^3}x\cos 3x + {\cos ^3}x\sin 3x = \frac{{3\sin x - \sin 3x}}{4}.\cos 3x + \frac{{3\cos x + \cos 3x}}{4}.\sin 3x$
$ = \frac{3}{4}\sin x\cos 3x - \frac{1}{4}\sin 3x\cos 3x + \frac{3}{4}\cos x\sin 3x + \frac{1}{4}\cos 3x\sin 3x$
$ = \frac{3}{8}\left[ {\sin \left( { - 2x} \right) + \sin 4x} \right] + \frac{3}{8}\left[ {\sin \left( { - 2x} \right) - \sin 4x} \right] = - \frac{3}{4}\sin 2x$
Vậy $I = - \frac{3}{4}\int {\sin 2xdx} = \frac{3}{8}\cos 2x + C$
$ \Rightarrow I = \int {\frac{{{t^2} + t}}{t}dt} = \int {tdt} + \int {\frac{{dt}}{t}} $
$ = \frac{1}{2}{t^2} + \ln \left| t \right| + C = \frac{1}{2}{\tan ^2}x + \ln \left| {\tan x} \right| + C$
$\Rightarrow I = \int {\frac{{dt}}{{{t^4}\left( {1 - {t^2}} \right)}}} = \int {\frac{{1 - {t^4} + {t^4}}}{{{t^4}\left( {1 - {t^2}} \right)}}dt} = \int {\frac{{1 + {t^2}}}{{{t^4}}}dt} + \int {\frac{{dt}}{{1 - {t^2}}}} $
$ = \int {\frac{{dt}}{{{t^4}}}} + \int {\frac{{dt}}{{{t^2}}}} - \int {\frac{{dt}}{{\left( {t - 1} \right)\left( {t + 1} \right)}}} = - \frac{1}{3}{t^{ - 3}} - \frac{1}{t} - \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right| + C$
$ = - \frac{1}{{3{{\sin }^3}x}} - \frac{1}{{\sin x}} - \frac{1}{2}\ln \left| {\frac{{\sin x - 1}}{{\sin x + 1}}} \right| + C$
$ = \frac{1}{4}\int {\left( {\sin 7x + \sin 5x} \right)dx} + \frac{1}{4}\int {\left( {\sin 3x + \sin x} \right)dx} $
$ = - \frac{1}{{28}}\cos 7x - \frac{1}{{20}}\cos 5x - \frac{1}{{12}}\cos 3x - \frac{1}{4}\cos x + C$
u = \frac{1}{{\sin x}}\\
dv = \frac{{dx}}{{{{\sin }^2}x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = - \frac{{\cos x}}{{{{\sin }^2}x}}dx\\
v = - \cot x
\end{array} \right.$
$ \Rightarrow I = - \frac{{\cot x}}{{\sin x}} - \int {\frac{{\cot x.\cos x}}{{{{\sin }^2}x}}dx} = - \frac{{\cot x}}{{\sin x}} - {I_1}$
Tính ${I_1} = \int {\frac{{{{\cos }^2}x}}{{{{\sin }^3}x}}dx} = \int {\frac{{1 - {{\sin }^2}x}}{{{{\sin }^3}x}}} dx = \int {\frac{{dx}}{{{{\sin }^3}x}} - \int {\frac{{dx}}{{\sin x}}} } = I - \ln \left| {\tan \frac{x}{2}} \right| + C$
$ \Rightarrow I = - \frac{{\cot x}}{{\sin x}} - {I_1} = - \frac{{\cot x}}{{\sin x}} - I + \ln \left| {\tan \frac{x}{2}} \right| + C$
$ \Rightarrow 2I = \ln \left| {\tan \frac{x}{2}} \right| - \frac{{\cot x}}{{\sin x}} + C \Rightarrow I = \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| - \frac{{\cot x}}{{2\sin x}} + C$
I. Dạng 1. $I = \int {\frac{{dx}}{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}} $
1. Phương pháp tính
Dùng đồng nhất thức:
$1 = \frac{{\sin \left( {a - b} \right)}}{{\sin \left( {a - b} \right)}} = \frac{{\sin \left[ {\left( {x + a} \right) - \left( {x + b} \right)} \right]}}{{\sin \left( {a - b} \right)}} = \frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\sin \left( {a - b} \right)}}$
Từ đó suy ra:
$I = \frac{1}{{\sin \left( {a - b} \right)}}\int {\frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}} dx$
$ = \frac{1}{{\sin \left( {a - b} \right)}}\int {\left[ {\frac{{\cos \left( {x + b} \right)}}{{\sin \left( {x + b} \right)}} - \frac{{\cos \left( {x + a} \right)}}{{\sin \left( {x + a} \right)}}} \right]} dx$
$ = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\ln \left| {\sin \left( {x + b} \right)} \right| - \ln \left| {\sin \left( {x + a} \right)} \right|} \right] + C$
2. Chú ý
Với cách này, ta có thể tìm được các nguyên hàm:
•$J = \int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} $ bằng cách dùng đồng nhất thức $1 = \frac{{\sin \left( {a - b} \right)}}{{\sin \left( {a - b} \right)}}$
•$K = \int {\frac{{dx}}{{\sin \left( {x + a} \right)\cos \left( {x + b} \right)}}} $ bằng cách dùng đồng nhất thức $1 = \frac{{\cos \left( {a - b} \right)}}{{\cos \left( {a - b} \right)}}$
3. Ví dụ áp dụng
• $I = \int {\frac{{dx}}{{\sin x\sin \left( {x + \frac{\pi }{6}} \right)}}} $
Ta có: $1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {x + \frac{\pi }{6}} \right) - x} \right]}}{{\frac{1}{2}}} = 2\left[ {\sin \left( {x + \frac{\pi }{6}} \right)\cos x - \cos \left( {x + \frac{\pi }{6}} \right)\sin x} \right]$
Từ đó: $I = 2\int {\frac{{\left[ {\sin \left( {x + \frac{\pi }{6}} \right)\cos x - \cos \left( {x + \frac{\pi }{6}} \right)\sin x} \right]}}{{\sin x\sin \left( {x + \frac{\pi }{6}} \right)}}} dx = 2\int {\left[ {\frac{{\cos x}}{{\sin x}} - \frac{{\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \right]} dx$
$ = 2\int {\frac{{d\left( {\sin x} \right)}}{{\sin x}}} - 2\int {\frac{{d\left( {\sin \left( {x + \frac{\pi }{6}} \right)} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = 2\ln \left| {\frac{{\sin x}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \right| + C$
- $I = \int {\frac{{dx}}{{\cos 3x\cos \left( {3x + \frac{\pi }{6}} \right)}}} $
$1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {3x + \frac{\pi }{6}} \right) - 3x} \right]}}{{\frac{1}{2}}} = 2\left[ {\sin \left( {3x + \frac{\pi }{6}} \right)\cos 3x - \cos \left( {3x + \frac{\pi }{6}} \right)\sin 3x} \right]$
Từ đó:
$I = 2\int {\frac{{\left[ {\sin \left( {3x + \frac{\pi }{6}} \right)\cos 3x - \cos \left( {3x + \frac{\pi }{6}} \right)\sin 3x} \right]}}{{\cos 3x\cos \left( {3x + \frac{\pi }{6}} \right)}}} dx = 2\int {\frac{{\sin \left( {3x + \frac{\pi }{6}} \right)}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}dx} - 2\int {\frac{{\sin 3x}}{{\cos 3x}}dx} $
$ = - \frac{2}{3}\int {\frac{{d\left( {\cos \left( {3x + \frac{\pi }{6}} \right)} \right)}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}} + \frac{2}{3}\int {\frac{{d\left( {\cos 3x} \right)}}{{\cos 3x}}} = \frac{2}{3}\ln \left| {\frac{{\cos 3x}}{{\cos \left( {3x + \frac{\pi }{6}} \right)}}} \right| + C$
- $I = \int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right)}}} $
$ = \sqrt 2 \left[ {\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{{12}}} \right)} \right]$
Từ đó: $I = \sqrt 2 \int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{{12}}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{{12}}} \right)}}} dx$
$ = \sqrt 2 \int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}dx} + \sqrt 2 \int {\frac{{\sin \left( {x + \frac{\pi }{{12}}} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} dx$
$ = \sqrt 2 \int {\frac{{d\left( {\sin \left( {x + \frac{\pi }{3}} \right)} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} - \sqrt 2 \int {\frac{{d\left( {\cos \left( {x + \frac{\pi }{{12}}} \right)} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} = \sqrt 2 \ln \left| {\frac{{\sin \left( {x + \frac{\pi }{3}} \right)}}{{\cos \left( {x + \frac{\pi }{{12}}} \right)}}} \right| + C$
II. Dạng 2. $I = \int {\tan \left( {x + a} \right)\tan \left( {x + b} \right)dx} $
1. Phương pháp tính
Ta có: $\tan \left( {x + a} \right)\tan \left( {x + b} \right) = \frac{{\sin \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}$
$ = \frac{{\sin \left( {x + a} \right)\sin \left( {x + b} \right) + \cos \left( {x + a} \right)\cos \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}} - 1 = \frac{{\cos \left( {a - b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}} - 1$
Từ đó: $I = \cos \left( {a - b} \right)\int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} - 1$
Đến đây ta gặp bài toán tìm nguyên hàm ở Dạng 1.
2. Chú ý
Với cách này, ta có thể tính được các nguyên hàm:
•$J = \int {\cot \left( {x + a} \right)\cot \left( {x + b} \right)dx} $
•$K = \int {\tan \left( {x + a} \right)\tan \left( {x + b} \right)dx} $
3. Ví dụ áp dụng
•$I = \int {\cot \left( {x + \frac{\pi }{3}} \right)\cot \left( {x + \frac{\pi }{6}} \right)dx} $
Ta có:
Ta có: $\cot \left( {x + \frac{\pi }{3}} \right)\cot \left( {x + \frac{\pi }{6}} \right) = \frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}$
$ = \frac{{\cos \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) + \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1$
$ = \frac{{\cos \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1 = \frac{{\sqrt 3 }}{2}.\frac{1}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} - 1$
Từ đó: $I = \frac{{\sqrt 3 }}{2}\int {\frac{1}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}dx} - \int {dx} = \frac{{\sqrt 3 }}{2}{I_1} - x + C$
Tính ${I_1} = \int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} $
Ta có: $1 = \frac{{\sin \frac{\pi }{6}}}{{\sin \frac{\pi }{6}}} = \frac{{\sin \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\frac{1}{2}}}$
$ = 2\left[ {\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)} \right]$
Từ đó: ${I_1} = 2\int {\frac{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} dx$
$ = 2\int {\frac{{\cos \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}dx} - 2\int {\frac{{\cos \left( {x + \frac{\pi }{3}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}dx} = 2\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| + C$
Suy ra: $I = \frac{{\sqrt 3 }}{2}.2\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| - x + C = \sqrt 3 \ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} \right| - x + C$
- $K = \int {\tan \left( {x + \frac{\pi }{3}} \right)\cot \left( {x + \frac{\pi }{6}} \right)} dx$
$ = \frac{{\sin \left( {x + \frac{\pi }{3}} \right)\cos \left( {x + \frac{\pi }{6}} \right) - \cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1$
$ = \frac{{\sin \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x + \frac{\pi }{6}} \right)} \right]}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1 = \frac{1}{2}.\frac{1}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}} + 1$
Từ đó: $K = \frac{1}{2}\int {\frac{1}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}dx} + \int {dx} = \frac{1}{2}{K_1} + x + C$
Đến đây, bằng cách tính ở Dạng 1, ta tính được:
${K_1} = \int {\frac{{dx}}{{\cos \left( {x + \frac{\pi }{3}} \right)\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{2}{{\sqrt 3 }}\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)}}} \right| + C$
Suy ra: $K = \frac{{\sqrt 3 }}{3}\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right)}}{{\cos \left( {x + \frac{\pi }{3}} \right)}}} \right| + x + C$
III. Dạng 3. $I = \int {\frac{{dx}}{{a\sin x + b\cos x}}} $
1. Phương pháp tính
Có: $a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \left( {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\sin x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\cos x} \right)$
$ \Rightarrow a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \sin \left( {x + \alpha } \right)$
$ \Rightarrow I = \frac{1}{{\sqrt {{a^2} + {b^2}} }}\int {\frac{{dx}}{{\sin \left( {x + \alpha } \right)}}} = \frac{1}{{\sqrt {{a^2} + {b^2}} }}\ln \left| {\tan \frac{{x + \alpha }}{2}} \right| + C$
2. Ví dụ áp dụng
- $I = \int {\frac{{2dx}}{{\sqrt 3 \sin x + \cos x}}} = \int {\frac{{dx}}{{\frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x}}} = \int {\frac{{dx}}{{\sin x\cos \frac{\pi }{6} + \cos x\sin \frac{\pi }{6}}}} $
- $J = \int {\frac{{dx}}{{\cos 2x - \sqrt 3 \sin 2x}}} = \frac{1}{2}\int {\frac{{dx}}{{\frac{1}{2}\cos 2x - \frac{{\sqrt 3 }}{2}\sin 2x}}} $
$ = - \frac{1}{4}\ln \left| {\tan \frac{{\frac{\pi }{6} - 2x}}{2}} \right| + C = - \frac{1}{4}\ln \left| {\tan \left( {\frac{\pi }{{12}} - x} \right)} \right| + C$
IV. Dạng 4. $I = \int {\frac{{dx}}{{a\sin x + b\cos x + c}}} $
1. Phương pháp tính
Đặt $\tan \frac{x}{2} = t \Rightarrow \left\{ \begin{array}{l}
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}\\
\tan x = \frac{{2t}}{{1 - {t^2}}}
\end{array} \right.$
2. Ví dụ áp dụng
- $I = \int {\frac{{dx}}{{3\cos x + 5\sin x + 3}}} $
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
Từ đó: $I = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{3.\frac{{1 - {t^2}}}{{1 + {t^2}}} + 5\frac{{2t}}{{1 + {t^2}}} + 3}}} = \int {\frac{{2dt}}{{3 - 3{t^2} + 10t + 3 + 3{t^2}}}} = \int {\frac{{2dt}}{{10t + 6}}} $
$ = \frac{1}{5}\int {\frac{{d\left( {5t + 3} \right)}}{{5t + 3}}} = \frac{1}{5}\ln \left| {5t + 3} \right| + C = \frac{1}{5}\ln \left| {5\tan \frac{x}{2} + 3} \right| + C$
- $J = \int {\frac{{2dx}}{{2\sin x - \cos x + 1}}} $
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
Từ đó: $J = \int {\frac{{2.\frac{{2dt}}{{1 + {t^2}}}}}{{2.\frac{{2t}}{{1 + {t^2}}} - \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{4dt}}{{4t - 1 + {t^2} + 1 + {t^2}}}} = \int {\frac{{4dt}}{{2{t^2} + 4t}}} = 2\int {\frac{{dt}}{{t\left( {t + 2} \right)}}} $
$ = \int {\left( {\frac{1}{t} - \frac{1}{{t + 2}}} \right)dt} = \ln \left| t \right| - \ln \left| {t + 2} \right| + C = \ln \left| {\tan \frac{x}{2}} \right| - \ln \left| {\tan \frac{x}{2} + 2} \right| + C$
- $K = \int {\frac{{dx}}{{\sin x + \tan x}}} $
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\tan x = \frac{{2t}}{{1 - {t^2}}}
\end{array} \right.$
Từ đó: $K = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{2t}}{{1 - {t^2}}}}}} = \frac{1}{2}\int {\frac{{1 - {t^2}}}{t}} dt = \frac{1}{2}\int {\frac{{dt}}{t}} - \frac{1}{2}\int {tdt} $
$ = \frac{1}{2}\ln \left| t \right| - \frac{1}{4}{t^2} + C = \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| - \frac{1}{4}{\tan ^2}\frac{x}{2} + C$
V. Dạng 5. $I = \int {\frac{{dx}}{{a.{{\sin }^2}x + b.\sin x\cos x + c.{{\cos }^2}x}}} $
1. Phương pháp tính
$I = \int {\frac{{dx}}{{\left( {a{{\tan }^2}x + b\tan x + c} \right).{{\cos }^2}x}}} $
Đặt $\tan x = t \Rightarrow \frac{{dx}}{{{{\cos }^2}x}} = dt$. Suy ra $I = \int {\frac{{dt}}{{a{t^2} + bt + c}}} $
2. Ví dụ áp dụng
- $I = \int {\frac{{dx}}{{3{{\sin }^2}x - 2\sin x\cos x - {{\cos }^2}x}}} = \int {\frac{{dx}}{{\left( {3{{\tan }^2}x - 2\tan x - 1} \right){{\cos }^2}x}}} $
$ \Rightarrow I = \int {\frac{{dt}}{{3{t^2} - 2t - 1}}} = \int {\frac{{dt}}{{\left( {t - 1} \right)\left( {3t + 1} \right)}}} $
$ = \frac{1}{4}\int {\left( {\frac{1}{{t - 1}} - \frac{3}{{3t + 1}}} \right)dt} = \frac{1}{4}\int {\frac{{dt}}{{t - 1}}} - \frac{1}{4}\int {\frac{{d\left( {3t + 1} \right)}}{{3t + 1}}} $
$ = \frac{1}{4}\ln \left| {\frac{{t - 1}}{{3t + 1}}} \right| + C = \frac{1}{4}\ln \left| {\frac{{\tan x - 1}}{{3\tan x + 1}}} \right| + C$
- $J = \int {\frac{{dx}}{{{{\sin }^2}x - 2\sin x\cos x - 2{{\cos }^2}x}}} = \int {\frac{{dx}}{{\left( {{{\tan }^2}x - 2\tan x - 2} \right){{\cos }^2}x}}} $
$ \Rightarrow J = \int {\frac{{dt}}{{{t^2} - 2t - 2}}} = \int {\frac{{d\left( {t - 1} \right)}}{{{{\left( {t - 1} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}} = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{t - 1 - \sqrt 3 }}{{t - 1 + \sqrt 3 }}} \right| + C$
$ = \frac{1}{{2\sqrt 3 }}\ln \left| {\frac{{\tan x - 1 - \sqrt 3 }}{{\tan x - 1 + \sqrt 3 }}} \right| + C$
VI. Dạng 6. $I = \int {\frac{{{a_1}\sin x + {b_1}\cos x}}{{{a_2}\sin x + {b_2}\cos x}}dx} $
1. Phương pháp tính
Ta tìm A, B sao cho:
${a_1}\sin x + {b_1}\cos x = A\left( {{a_2}\sin x + {b_2}\cos x} \right) + B\left( {{a_2}\cos x - {b_2}\sin x} \right)$
2. Ví dụ áp dụng
- $I = \int {\frac{{4\sin x + 3\cos x}}{{\sin x + 2\cos x}}dx} $
$4\sin x + 3\cos x = A\left( {\sin x + 2\cos x} \right) + B\left( {\cos x - 2\sin x} \right)$
$ \Rightarrow 4\sin x + 3\cos x = \left( {A - 2B} \right)\sin x + \left( {2A + B} \right)\cos x \Rightarrow \left\{ \begin{array}{l}
A - 2B = 4\\
2A + B = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 2\\
B = - 1
\end{array} \right.$
Từ đó: $I = \int {\frac{{2\left( {\sin x + 2\cos x} \right) - \left( {\cos x - 2\sin x} \right)}}{{\sin x + 2\cos x}}} dx$
$ = 2\int {dx} - \int {\frac{{d\left( {\sin x + 2\cos x} \right)}}{{\sin x + 2\cos x}}} = 2x - \ln \left| {\sin x + 2\cos x} \right| + C$
- $J = \int {\frac{{3\cos x - 2\sin x}}{{\cos x - 4\sin x}}} dx$
$3\cos x - 2\sin x = A\left( {\cos x - 4\sin x} \right) + B\left( { - \sin x - 4\cos x} \right)$
$ \Rightarrow 3\cos x - 2\sin x = \left( {A - 4B} \right)\cos x + \left( { - 4A - B} \right)\sin x$
$ \Rightarrow \left\{ \begin{array}{l}
A - 4B = 3\\
4A + B = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = \frac{{11}}{{17}}\\
B = - \frac{{10}}{{17}}
\end{array} \right.$
Từ đó: $J = \int {\frac{{\frac{{11}}{{17}}\left( {\cos x - 4\sin x} \right) - \frac{{10}}{{17}}\left( { - \sin x - 4\cos x} \right)}}{{\cos x - 4\sin x}}} dx$
$ = \frac{{11}}{{17}}\int {dx} - \frac{{10}}{{17}}\int {\frac{{d\left( {\cos x - 4\sin x} \right)}}{{\cos x - 4\sin x}}} = \frac{{11}}{{17}}x - \frac{{10}}{{17}}\ln \left| {\cos x - 4\sin x} \right| + C$
3. Chú ý
- Nếu gặp $I = \int {\frac{{{a_1}\sin x + {b_1}\cos x}}{{{{\left( {{a_2}\sin x + {b_2}\cos x} \right)}^2}}}} dx$ ta vẫn tìm A, B sao cho: ${a_1}\sin x + {b_1}\cos x = A\left( {{a_2}\sin x + {b_2}\cos x} \right) + B\left( {{a_2}\cos x - {b_2}\sin x} \right)$
- Nếu gặp $I = \int {\frac{{{a_1}\sin x + {b_1}\cos x + {c_1}}}{{{a_2}\sin x + {b_2}\cos x + {c_2}}}} dx$ ta tìm A, B sao cho:
Chẳng hạn:
- $I = \int {\frac{{8\cos x}}{{{{\left( {\sqrt 3 \sin x + \cos x} \right)}^2}}}dx} $
$8\cos x = A\left( {\sqrt 3 \sin x + \cos x} \right) + B\left( {\sqrt 3 \cos x - \sin x} \right)$
$ \Rightarrow 8\cos x = \left( {A\sqrt 3 - B} \right)\sin x + \left( {A + B\sqrt 3 } \right)\cos x$
$ \Rightarrow \left\{ \begin{array}{l}
A\sqrt 3 - B = 0\\
A + B\sqrt 3 = 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 2\\
B = 2\sqrt 3
\end{array} \right.$
Từ đó: $I = \int {\frac{{2\left( {\sqrt 3 \sin x + \cos x} \right) + 2\sqrt 3 \left( {\sqrt 3 \cos x - \sin x} \right)}}{{{{\left( {\sqrt 3 \sin x + \cos x} \right)}^2}}}} dx$
$ = 2\int {\frac{{dx}}{{\sqrt 3 \sin x + \cos x}}} + 2\sqrt 3 \int {\frac{{d\left( {\sqrt 3 \sin x + \cos x} \right)}}{{{{\left( {\sqrt 3 \sin x + \cos x} \right)}^2}}}} = 2{I_1} - \frac{{2\sqrt 3 }}{{\sqrt 3 \sin x + \cos x}} + C$
Tìm ${I_1} = \int {\frac{{dx}}{{\sqrt 3 \sin x + \cos x}}} = \frac{1}{2}\int {\frac{{dx}}{{\frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x}}} = \frac{1}{2}\int {\frac{{dx}}{{\sin x\cos \frac{\pi }{6} + \cos x\sin \frac{\pi }{6}}}} $
$ = \frac{1}{2}\int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{1}{2}\int {\frac{{d\left( {x + \frac{\pi }{6}} \right)}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} = \frac{1}{2}\ln \left| {\tan \frac{{x + \frac{\pi }{6}}}{2}} \right| + C = \frac{1}{2}\ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right)} \right| + C$
Vậy $I = \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right)} \right| - \frac{{2\sqrt 3 }}{{\sqrt 3 \sin x + \cos x}} + C$
- $J = \int {\frac{{8\sin x + \cos x + 5}}{{2\sin x - \cos x + 1}}dx} $
$8\sin x + \cos x + 5 = A\left( {2\sin x - \cos x + 1} \right) + B\left( {2\cos x + \sin x} \right) + C$
$ \Rightarrow 8\sin x + \cos x + 5 = \left( {2A + B} \right)\sin x + \left( { - A + 2B} \right)\cos x + A + C$
$ \Rightarrow \left\{ \begin{array}{l}
2A + B = 8\\
- A + 2B = 1\\
A + C = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 3\\
B = 2\\
C = 2
\end{array} \right.$
Từ đó: $J = \int {\frac{{3\left( {2\sin x - \cos x + 1} \right) + 2\left( {2\cos x + \sin x} \right) + 2}}{{2\sin x - \cos x + 1}}} dx$
$ = 3\int {dx} + 2\int {\frac{{2\cos x + \sin x}}{{2\sin x - \cos x + 1}}} dx + 2\frac{{dx}}{{2\sin x - \cos x + 1}}$
$ = 3x + 2\ln \left| {2\sin x - \cos x + 1} \right| + 2{J_1}$
Tìm ${J_1} = \int {\frac{{dx}}{{2\sin x - \cos x + 1}}} $
Đặt $\tan \frac{x}{2} = t \Rightarrow \left\{ \begin{array}{l}
dx = \frac{{2dt}}{{1 + {t^2}}}\\
\sin x = \frac{{2t}}{{1 + {t^2}}}\\
\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}
\end{array} \right.$
$ \Rightarrow {J_1} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{2.\frac{{2t}}{{1 + {t^2}}} - \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{dt}}{{{t^2} + 2t}}} = \int {\frac{{dt}}{{t\left( {t + 2} \right)}}} = \frac{1}{2}\int {\left( {\frac{1}{t} - \frac{1}{{t + 2}}} \right)} dt$
$ = \frac{1}{2}\ln \left| {\frac{t}{{t + 2}}} \right| + C = \frac{1}{2}\ln \left| {\frac{{\tan \frac{x}{2}}}{{\tan \frac{x}{2} + 2}}} \right| + C$
Vậy: $J = 3x + 2\ln \left| {2\sin x - \cos x + 1} \right| + \ln \left| {\frac{{\tan \frac{x}{2}}}{{\tan \frac{x}{2} + 2}}} \right| + C$
VII. Dạng 7. Biến đổi đưa về nguyên hàm cơ bản hoặc 6 dạng ở trên
- $I = \int {\cos 3x\cos 4xdx} = \frac{1}{2}\int {\left( {\cos x + \cos 7x} \right)dx} $
- • $I = \int {\cos x\sin 2x\cos 3xdx} = \frac{1}{2}\int {\sin 2x\left( {\cos 2x + \cos 4x} \right)} dx$
$ = \frac{1}{4}\int {\sin 2xd\left( {\sin 2x} \right)} + \frac{1}{4}\int {\left( { - \sin 2x + \sin 6x} \right)} dx$
$ = \frac{1}{8}{\sin ^2}2x + \frac{1}{8}\cos 2x - \frac{1}{{24}}\cos 6x + C$
- $I = \int {\tan x\tan \left( {\frac{\pi }{3} - x} \right)\tan \left( {\frac{\pi }{3} + x} \right)dx} $
$ = \frac{{\sin x\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right)}}{{\cos x\left( {\cos 2x + \cos \frac{{2\pi }}{3}} \right)}} = \frac{{\sin x\left( {1 - 2{{\sin }^2}x + \frac{1}{2}} \right)}}{{\cos x\left( {2{{\cos }^2}x - 1 - \frac{1}{2}} \right)}}$
$ = \frac{{\sin x\left( {3 - 4{{\sin }^2}x} \right)}}{{\cos x\left( {4{{\cos }^2}x - 3} \right)}} = \frac{{3\sin x - 4{{\sin }^3}x}}{{4{{\cos }^3}x - 3\cos x}} = \frac{{\sin 3x}}{{\cos 3x}}$
Từ đó: $I = \int {\frac{{\sin 3x}}{{\cos 3x}}dx} = - \frac{1}{3}\int {\frac{{d\left( {\cos 3x} \right)}}{{\cos 3x}}} = - \frac{1}{3}\ln \left| {\cos 3x} \right| + C$
- $I = \int {{{\sin }^3}x\sin 3xdx} $
$ \Rightarrow {\sin ^3}x\sin 3x = \frac{{3\sin x - 4\sin 3x}}{4}.\sin 3x$
$ = \frac{3}{4}\sin x\sin 3x - \frac{1}{4}{\sin ^2}3x = \frac{3}{8}\left( {\cos 2x - \cos 4x} \right) - \frac{1}{8}\left( {1 - \cos 6x} \right)$
$ = \frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x - \frac{1}{8}$
Từ đó: $I = \int {\left( {\frac{3}{8}\cos 2x - \frac{3}{8}\cos 4x + \frac{1}{8}\cos 6x - \frac{1}{8}} \right)} dx$
$ = \frac{3}{{16}}\sin 2x - \frac{3}{{32}}\sin 4x + \frac{1}{{48}}\sin 6x - \frac{1}{8}x + C$
- $I = \int {\left( {{{\sin }^3}x\cos 3x + {{\cos }^3}x\sin 3x} \right)dx} $
${\cos ^3}x = \frac{{3\cos x + \cos 3x}}{4}$
Suy ra: ${\sin ^3}x\cos 3x + {\cos ^3}x\sin 3x = \frac{{3\sin x - \sin 3x}}{4}.\cos 3x + \frac{{3\cos x + \cos 3x}}{4}.\sin 3x$
$ = \frac{3}{4}\sin x\cos 3x - \frac{1}{4}\sin 3x\cos 3x + \frac{3}{4}\cos x\sin 3x + \frac{1}{4}\cos 3x\sin 3x$
$ = \frac{3}{8}\left[ {\sin \left( { - 2x} \right) + \sin 4x} \right] + \frac{3}{8}\left[ {\sin \left( { - 2x} \right) - \sin 4x} \right] = - \frac{3}{4}\sin 2x$
Vậy $I = - \frac{3}{4}\int {\sin 2xdx} = \frac{3}{8}\cos 2x + C$
- $I = \int {\frac{{dx}}{{\sin x{{\cos }^3}x}}} = \int {\frac{{dx}}{{\tan x{{\cos }^4}x}}} = \int {\frac{1}{{\tan x}}.\frac{1}{{{{\cos }^2}x}}.\frac{{dx}}{{{{\cos }^2}x}}} = \int {\frac{1}{{\tan x}}\left( {1 + {{\tan }^2}x} \right)\frac{{dx}}{{{{\cos }^2}x}}} $
$ \Rightarrow I = \int {\frac{{{t^2} + t}}{t}dt} = \int {tdt} + \int {\frac{{dt}}{t}} $
$ = \frac{1}{2}{t^2} + \ln \left| t \right| + C = \frac{1}{2}{\tan ^2}x + \ln \left| {\tan x} \right| + C$
- $I = \int {\frac{{dx}}{{{{\sin }^4}x\cos x}}} = \int {\frac{{\cos xdx}}{{{{\sin }^4}x{{\cos }^2}x}}} $
$\Rightarrow I = \int {\frac{{dt}}{{{t^4}\left( {1 - {t^2}} \right)}}} = \int {\frac{{1 - {t^4} + {t^4}}}{{{t^4}\left( {1 - {t^2}} \right)}}dt} = \int {\frac{{1 + {t^2}}}{{{t^4}}}dt} + \int {\frac{{dt}}{{1 - {t^2}}}} $
$ = \int {\frac{{dt}}{{{t^4}}}} + \int {\frac{{dt}}{{{t^2}}}} - \int {\frac{{dt}}{{\left( {t - 1} \right)\left( {t + 1} \right)}}} = - \frac{1}{3}{t^{ - 3}} - \frac{1}{t} - \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right| + C$
$ = - \frac{1}{{3{{\sin }^3}x}} - \frac{1}{{\sin x}} - \frac{1}{2}\ln \left| {\frac{{\sin x - 1}}{{\sin x + 1}}} \right| + C$
- $I = \int {\frac{{\sin 3x\sin 4x}}{{\tan x + \tan 2x}}dx} = \int {\frac{{\sin 3x\sin 4x}}{{\frac{{\sin 3x}}{{\cos x\cos 2x}}}}dx} = \int {\sin 4x\cos 2x\cos xdx} $
$ = \frac{1}{4}\int {\left( {\sin 7x + \sin 5x} \right)dx} + \frac{1}{4}\int {\left( {\sin 3x + \sin x} \right)dx} $
$ = - \frac{1}{{28}}\cos 7x - \frac{1}{{20}}\cos 5x - \frac{1}{{12}}\cos 3x - \frac{1}{4}\cos x + C$
- $I = \int {\frac{{dx}}{{{{\sin }^3}x}}} $
u = \frac{1}{{\sin x}}\\
dv = \frac{{dx}}{{{{\sin }^2}x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = - \frac{{\cos x}}{{{{\sin }^2}x}}dx\\
v = - \cot x
\end{array} \right.$
$ \Rightarrow I = - \frac{{\cot x}}{{\sin x}} - \int {\frac{{\cot x.\cos x}}{{{{\sin }^2}x}}dx} = - \frac{{\cot x}}{{\sin x}} - {I_1}$
Tính ${I_1} = \int {\frac{{{{\cos }^2}x}}{{{{\sin }^3}x}}dx} = \int {\frac{{1 - {{\sin }^2}x}}{{{{\sin }^3}x}}} dx = \int {\frac{{dx}}{{{{\sin }^3}x}} - \int {\frac{{dx}}{{\sin x}}} } = I - \ln \left| {\tan \frac{x}{2}} \right| + C$
$ \Rightarrow I = - \frac{{\cot x}}{{\sin x}} - {I_1} = - \frac{{\cot x}}{{\sin x}} - I + \ln \left| {\tan \frac{x}{2}} \right| + C$
$ \Rightarrow 2I = \ln \left| {\tan \frac{x}{2}} \right| - \frac{{\cot x}}{{\sin x}} + C \Rightarrow I = \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| - \frac{{\cot x}}{{2\sin x}} + C$
- $I = \int {\frac{{dx}}{{{{\cos }^3}x}} = } ... = \frac{1}{3}\left[ {\frac{{\tan x}}{{\cos x}} + \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|} \right] + C$
