Question

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

• A. $\frac{1}{2}\log 2$
• B. $\log 2$
• C. $\frac{1}{2}\log 3$
• D. $\log 3$
• E. $\frac{1}{3}\log 3$

Answer 1

Answer: (C)

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

Let $l={\int }_0^{\pi /2}\frac{\sin 2x}{1+2{\cos }^{2}x}dx$
Let $t=1+2{\cos }^{2}x$
$\Rightarrow dt=-4\cos x\sin xdx=-2\sin 2xdx$
Then, $l={\int }_3^1\frac{dt}{-2t}=\frac{1}{2}{\int }_1^3\frac{dt}{t}$
$=\frac{1}{2}[\log |t|{]}_1^3$
$=\frac{1}{2}(\log 3-\log 1)$
$=\frac{1}{2}\log 3$