Q112 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
$\int_0^{\frac{\pi}{4}} \frac{\sec x}{1+2 \sin ^2 x} d x=$
- Correct answerA$\frac{1}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12}$
- B$\frac{2}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{6}$
- C$\frac{1}{6} \log (\sqrt{2}-1)+\frac{\pi}{12}$
- D$\frac{1}{4} \log (\sqrt{2}-1)-\frac{\pi \sqrt{3}}{6}$
Solution
$$
\begin{aligned}
& \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x \\
= & \int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x\left(1+2 \sin ^2 x\right)} d x \\
= & \int_0^{\pi / 4} \frac{\cos x d x}{\left(1-\sin ^2 x\right)\left(1+2 \sin ^2 x\right)} d x
\end{aligned}
$$
Let $\sin x=t \therefore \sin 0=0$
$$
\begin{aligned}
& \cos x d x=d t ; \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\
& \int_0^{1 / \sqrt{2}} \frac{d t}{\left(1-t^2\right)\left(1+2 t^2\right)}=\int_0^{1 / \sqrt{2}}\left(\frac{1}{1-t^2}+\frac{2}{1+2 t^2}\right) d t \\
& =\frac{1}{3}\left[\frac{1}{2} \log \left(\frac{1+t}{1-t}\right)+\frac{2}{\sqrt{2}} \tan ^{-1} \sqrt{2} t\right]_0^{1 / \sqrt{2}}
\end{aligned}
$$
$\begin{aligned} & =\frac{1}{3}\left[\frac{1}{2} \log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)+\sqrt{2} \tan ^{-1} 1\right] \\ & =\frac{1}{3}\left[\frac{1}{2} \log (\sqrt{2}+1)^2+\sqrt{2} \cdot \frac{\pi}{4}\right] \\ & =\frac{1}{3}\left[\log (\sqrt{2}+1)+\frac{\pi}{2 \sqrt{2}}\right]=\frac{1}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12} .\end{aligned}$
Hence, the correct answer is option: 'A'.