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Question Number 61522 by YSN 1905 last updated on 03/Jun/19

$$I=\int \frac{\sin x.e^{\cos x}-(\sin x+\cos x)e^{(\sin x+\cos x)}}{e^{2\sin x}-2e^{\sin x}+1}dx$$

Answered by perlman last updated on 04/Jun/19

$$I=\int \frac{sin\left(x\right)e^{cos\left(x\right)}-(sin\left(x\right)+cos\left(x\right))e^{sin\left(x\right)+cos\left(x\right)}}{{(e^{sin\left(x\right)}-1)}^2}$$$$=\int \frac{-sin\left(x\right)e^{cosx}(e^{sin\left(x\right)}-1)-cos\left(x\right)e^{sin\left(x\right)}.e^{cos\left(x\right)}}{{(e^{sin\left(x\right)}-1)}^2}$$$$=\int \frac{d\left(e^{cos\left(x\right)}\right)(sin\left(x\right)-1)-d(sin\left(x\right)-1)e^{cos\left(x\right)}}{(e^{sin\left(x\right)}-1)2}$$$$=\int d\frac{e^{cos\left(x\right)}}{(e^{sin\left(x\right)}-1)}=\frac{e^{cos\left(x\right)}}{e^{sin\left(x\right)}-1}+c$$

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