Question

$\int \frac {\sin \,2x}{\sin^2x +2 \,\cos^2x}dx$ =

• A. $\log (1 + \cos^2x) + C$
• B. $\log (1 + \tan^2x) + C$
• C. $- \log(1+\sin^2x) + C$
• D. $- \log (1 + \cos^2x) + C$

Answer 1

Answer: (D)

Let $I =\int \frac{\sin 2 x}{\sin ^{2} x+2 \cos ^{2} x} d x$
$=\int \frac{\sin 2 x}{1-\cos ^{2} x+2 \cos ^{2} x} d x $
$=\int \frac{\sin 2 x}{1+\cos ^{2} x} d x$
Put $1+\cos ^{2} x=t$
$\Rightarrow -2 \cos x \sin x d x=d t$
$\Rightarrow -\sin 2 x d x=d t$
$\therefore I =-\int \frac{d t}{t}$
$=-\log t +C$
$=-\log \left(1+\cos ^{2} x\right)+C$