Question

Integrate the following function: $\frac{\sin x}{(1+\cos x{)}^2}$

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

Answer 1

The correct option is D $\frac{1}{1+\cos x}+c$

Integrated the following function $\int \frac{\sin x}{(1+\cos n{)}^2}dx$

Let $1+\cos x=t$

$-\sin dx=dt$

$dx=\frac{dt}{-\sin x}$

$I=\int \frac{\sin x}{t^2}\frac{dt}{-\sin x}$

$I=-\int \frac{dt}{t^2}$ $\left[\int x^{n}dn=\frac{x^{n+1}}{n+1}\right]$

$I=\frac{1}{t}+C$

$I=\frac{1}{(1+\cos x)+C}$

So, option D is correct