Question

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

• A. $\log (1+co{s}^{2}x)+C$
• B. $\log (1+ta{n}^{2}x)+C$
• C. $-\log (1+si{n}^{2}x)+C$
• D. $-\log (1+co{s}^{2}x)+C$

Answer 1

The correct option is D $-\log (1+co{s}^{2}x)+C$

Let $I=\int \frac{\sin 2x}{{\sin }^{2}x+2{\cos }^{2}x}dx=\int \frac{\sin 2x}{({\sin }^{2}x+{\cos }^{2}x)+{\cos }^{2}x}dx$

$=\int \frac{\sin 2x}{1+{\cos }^{2}x}dx$

Substitute $1+{\cos }^{2}x=u$

$\Rightarrow 0+2\cos x(-\sin x)dx=du$

$\Rightarrow \sin 2xdx=-du$

Therefore $I=-\int \frac{du}{u}=-\ln u+c=-\ln (1+{\cos }^{2}x)+c$