If $\int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log _{\mathrm{e}}\left(\frac{\mathrm{a}}{3}\right)+\frac{\pi}{\mathrm{b} \sqrt{3}}$, where $\mathrm{a}, \mathrm{b} \in \mathbf{N}$, then $\mathrm{a}+\mathrm{b}$ is equal to_________

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Aicharya · Accepted Answer

$\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\frac{1}{2} \sin 2 x} d x=\int_0^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2+\sin 2 x} d x \\ & \int \frac{1}{2+\sin 2 x}-\int \frac{\cos 2 x}{2+\sin 2 x} \end{aligned}$ $\left(\mathrm{I}_1\right) \quad-\quad\left(\mathrm{I}_2\right)$
$\begin{aligned} & \left(\mathrm{I}_1\right)=\int \frac{\mathrm{dx}}{2+\frac{2 \tan \mathrm{x}}{1+\tan ^2 x}} \\ & \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \mathrm{xdx}}{2 \tan ^2 \mathrm{x}+2 \tan x+2} \end{aligned}$ $\begin{aligned} & \tan x=t \\ & \frac{1}{2} \int_0^1 \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{\pi}{6 \sqrt{3}} \\ & \mathrm{I}_2=\int_0^{\pi / 4} \frac{\cos 2 \mathrm{x}}{2+\sin 2 \mathrm{x}} \mathrm{dx}=\frac{1}{2}\left(\ln \frac{3}{2}\right) \\ & \mathrm{I}_1-\mathrm{I}_2=\frac{1}{\sqrt{3}} \frac{\pi}{6}+\frac{1}{2} \ln \frac{2}{3} \\ & \Rightarrow \mathrm{a}=2, \mathrm{~b}=6 \end{aligned}$
Ans. 8

If $\int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log…

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