tính tích phân

T

tuyn

\int_{}^{}tanxtan(x+pi/6)dx
giúp mình với cám ơn nhiều
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H

hoanghondo94

[TEX] I= \int tanxtan\left ( x+\frac{\pi }{6} \right )dx[/TEX]

[TEX] \left\{ \begin{array}{l} u=tan(x+\frac{\pi }{6}) \\ dv=tanxdx \end{array} \right [/TEX]
[TEX]\Rightarrow \left\{ \begin{array}{l} du=\frac{1}{cos^2(x+\frac{\pi }{6})}dx \\ v=-ln|cosx| \end{array} \right[/TEX]

[TEX] I=-tan(x+\frac{\pi }{6})ln|cosx|+\int ln|cosx| \frac{1}{cos^2(x+\frac{\pi }{6})}dx[/TEX]
[TEX]=-tan(x+\frac{\pi }{6})ln|cosx|+I'+C[/TEX]

[TEX]I'=ln|cosx|.tan(x+\frac{\pi }{6})+\int tanx.tan(x+\frac{\pi }{6})dx[/TEX]

[TEX]=ln|cosx|.tan(x+\frac{\pi }{6})+I[/TEX]
 
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