Question

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

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

Answer 1

Answer: (D)

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