Question

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

• A. $\frac{e^{x} \cos x}{1+\sin x}+C$
• B. $C-\frac{e^{x} \sin x}{1+\sin x}$
• C. $C-\frac{e^{x}}{1+\sin x}$
• D. $C-\frac{e^{x} \cos x}{1+\sin x}$

Answer 1

Answer: (A)

$I=\int\left(\frac{\cos x-1}{\sin x+1}\right) e^{x} d x$
$I=\int e^{x}\left(\frac{\cos x}{\sin x+1}-\frac{1}{1+\sin x}\right) d x$
Let $f(x)=\frac{\cos x}{1+\sin x}$
$f'(x)=\frac{-\sin x(1+\sin x)-\cos ^{2} x}{(1+\sin x)^{2}}$
$=\frac{-1-\sin x}{(1+\sin x)^{2}}=-\frac{1}{1+\sin x}$
$I=\int e^{x}\left(f(x)+f'(x)\right] d x$
$I=e^{x} \cdot f(x)=e^{x} \frac{\cos x}{1+\sin x}$
$I=\frac{\cos x \cdot e^{x}}{1+\sin x}+C$