Question

Tính: \(\int\dfrac{ln\left(sinx+cosx\right)}{cos^2x}dx\)

Answer 1

Hi

Answer 2

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Answer 3

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.chào cậu :)))))))))))))

Answer 4

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Answer 5

Đặt \(\left\{{}\begin{matrix}u=ln\left(sinx+cosx\right)\\dv=\dfrac{dx}{cos^2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{cosx-sinx}{sinx+cosx}dx\\v=tanx\end{matrix}\right.\)

\(I=tanx.ln\left(sinx+cosx\right)-\int\dfrac{\left(cosx-sinx\right)tanx}{\left(sinx+cosx\right)}dx\)

Xét \(J=\int\dfrac{\left(cosx-sinx\right)tanx}{sinx+cosx}dx=\int\left(\dfrac{sinx-cosx}{sinx+cosx}+1-tanx\right)dx\)

\(=-ln\left|sinx+cosx\right|+x+ln\left|cosx\right|+C\)

Vậy \(I=tanx.ln\left|sinx+cosx\right|+ln\left|sinx+cosx\right|-x-ln\left|cosx\right|+C\)