Question

If $\underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{\cos x}{3+4\sin x}dx=k\log \left(\frac{3+2\sqrt{3}}{3}\right)$ then $k$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

We have,
$\underset{0}{\overset{\pi /3}{\int }}\frac{\cos x}{3+4\sin x}dx$
$=k\log \left(\frac{3+2\sqrt{3}}{3}\right)\dots \left(i\right)$
Let $I=\underset{0}{\overset{\pi /3}{\int }}\frac{\cos x}{3+4\sin x}dx$
Put $3+4\sin x=t$
$\Rightarrow 0+4\cos xdx=dt$
Upper limit, $x=\frac{\pi }{3},t=3+4\sin \frac{\pi }{3}$
$=3+4\times \frac{\sqrt{3}}{2}=3+2\sqrt{3}$
and lower limit $x=0$,
$t=3+4\sin 0=3$
$\therefore I=\underset{3}{\overset{3+2\sqrt{3}}{\int }}\frac{dt}{4t}=\frac{1}{4}{\left[\log t\right]}_3^{3+2\sqrt{3}}$
$=\frac{1}{4}\left[\log \left(3+2\sqrt{3}\right)-\log 3\right]$
$\Rightarrow I=\frac{1}{4}\log \left(\frac{3+2\sqrt{3}}{3}\right)\dots \left(ii\right)$
$\therefore$ From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$k=\frac{1}{4}$