Question

If $\int \frac{\cos x-1}{\sin x+1}{e}^{x}dx$ 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)

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