Question

${\int }_0^{\pi /2}\frac{\cos x}{(1+\sin x)(2+\sin x)}dx$ is equal to

• A. $\ell$ $n\frac{3}{4}$
• B. $\ell$ $n\frac{2}{3}$
• C. $\ell$ $n\frac{4}{3}$
• D. 0

Answer 1

The correct option is C $\ell$ $n\frac{4}{3}$

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{cosx}{(1+sinx)(2+sinx)}\right)dx$

$letsinx=t$ $thencosxdx=dt$

$alsoifx=0$ $thent=sin0=0$

$andifx=\left(\frac{\pi }{2}\right)thent=1$

$\therefore I={\int }_0^1\frac{1}{(1+t)(2+t)}dt$

$={\int }_0^1[\frac{1}{1+t}-\frac{1}{2+t}]dt$

$={\left[log\right(1+t)-log(2+t\left)\right]}_0^1$

$={\left[log\left(\frac{1+t}{2+t}\right)\right]}_0^1$

$=log\frac{2}{3}-log\frac{1}{2}$

$=log\frac{4}{3}$