Q41 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
$\int \frac{\sin (2 x)}{\sin ^2(x)+2 \cos ^2(x)} d x=$
- A$\log \left|1+\cos ^2(x)\right|+c$
- B$-\log \left|1+\sin ^2(x)\right|+c$
- C$\log \left|1+\tan ^2(x)\right|+c$
- Correct answerD$-\log \left|1+\cos ^2(x)\right|+c$
Solution
$$
\begin{aligned}
& \text { } \int \frac{\sin 2 x}{\sin ^2 x+2 \cos ^2 x} d x \\
& \quad=\int \frac{\sin 2 x}{1-\cos ^2 x+2 \cos ^2 x} d x=\int \frac{\sin 2 x}{1+\cos ^2 x} d x
\end{aligned}
$$
Put, $\cos ^2 x=t$
$$
\begin{aligned}
2 \cos x( & -\sin x) d x=d t \\
- & \sin 2 x d x=d t \\
& =\int \frac{-d t}{1+t}=-\log |1+t|+c \quad\left[\because t=\cos ^2 x\right] \\
& =-\log \left|1+\cos ^2 x\right|+c
\end{aligned}
$$
Hence, option (d) is correct.
Hence, the correct answer is option: 'D'.