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Question

Mathematics Question on Definite Integral

If cosx1sinx+1exdx\int \frac{\cos x-1}{\sin x+1} e^{x} d x is equal to:

A

excosx1+sinx+C\frac{e^{x} \cos x}{1+\sin x}+C

B

Cexsinx1+sinxC-\frac{e^{x} \sin x}{1+\sin x}

C

Cex1+sinxC-\frac{e^{x}}{1+\sin x}

D

Cexcosx1+sinxC-\frac{e^{x} \cos x}{1+\sin x}

Answer

excosx1+sinx+C\frac{e^{x} \cos x}{1+\sin x}+C

Explanation

Solution

I=(cosx1sinx+1)exdxI=\int\left(\frac{\cos x-1}{\sin x+1}\right) e^{x} d x
I=ex(cosxsinx+111+sinx)dxI=\int e^{x}\left(\frac{\cos x}{\sin x+1}-\frac{1}{1+\sin x}\right) d x
Let f(x)=cosx1+sinxf(x)=\frac{\cos x}{1+\sin x}
f(x)=sinx(1+sinx)cos2x(1+sinx)2f'(x)=\frac{-\sin x(1+\sin x)-\cos ^{2} x}{(1+\sin x)^{2}}
=1sinx(1+sinx)2=11+sinx=\frac{-1-\sin x}{(1+\sin x)^{2}}=-\frac{1}{1+\sin x}
I=ex(f(x)+f(x)]dxI=\int e^{x}\left(f(x)+f'(x)\right] d x
I=exf(x)=excosx1+sinxI=e^{x} \cdot f(x)=e^{x} \frac{\cos x}{1+\sin x}
I=cosxex1+sinx+CI=\frac{\cos x \cdot e^{x}}{1+\sin x}+C