_0^ / 4 x 1+2 ^2 x d x is equal to - Vidyadip

MCQ

$\int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x$ is equal to

✓
$\frac{1}{3}\left[\log (\sqrt{2}+1)+\frac{\pi}{2 \sqrt{2}}\right]$
B
$\frac{1}{3}\left[\log (\sqrt{2}+1)-\frac{\pi}{2 \sqrt{2}}\right]$
C
$3\left[\log (\sqrt{2}+1)-\frac{\pi}{2 \sqrt{2}}\right]$
D
$3\left[\log (\sqrt{2}+1)+\frac{\pi}{2 \sqrt{2}}\right]$

✓

Answer

Correct option: A.

$\frac{1}{3}\left[\log (\sqrt{2}+1)+\frac{\pi}{2 \sqrt{2}}\right]$

(A)
Let $I =\int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x$
$\begin{array}{l}=\int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x\left(1+2 \sin ^2 x\right)} d x \\ =\int_0^{\pi / 4} \frac{\cos x}{\left(1-\sin ^2 x\right)\left(1+2 \sin ^2 x\right)} d x\end{array}$
Put $\sin x= t \Rightarrow \cos x d x= dt$
$\therefore \quad I=\int_0^{1 / \sqrt{2}} \frac{1}{\left(1-t^2\right)\left(1+2 t^2\right)} d t$
$\begin{array}{l}=\frac{1}{3} \int_0^{1 / \sqrt{2}}\left(\frac{1}{1- t ^2}+\frac{2}{1+2 t ^2}\right) dt \\ =\frac{1}{3}\left[\frac{1}{2 \cdot 1} \log \left(\frac{1+ t }{1- t }\right)+\frac{2}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t )\right]_0^{\frac{1}{\sqrt{2}}} \\ =\frac{1}{3}\left[\frac{1}{2} \log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)+\sqrt{2} \tan ^{-1} 1\right] \\ =\frac{1}{3}\left[\frac{1}{2} \log (\sqrt{2}+1)^2+\sqrt{2} \cdot \frac{\pi}{4}\right] \\ =\frac{1}{3}\left[\log (\sqrt{2}+1)+\frac{\pi}{2 \sqrt{2}}\right]\end{array}$

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