pi}{4}} \frac{\sin ^2 x}{\sin ^2 x+\cos ^2 x+\sin x \cos x} d x$$

$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\tan ^2 x}{1+\tan x+\tan ^2 x} d x$$

$$=\int_\limits0^{\frac{\pi}{4}} \frac{\tan x \cdot \sec ^2 x d x}{\left(1+\tan ^2 x\right)\left(1+\tan x+\tan ^2 x\right)}$$

Let $$\tan x=t$$

$$\begin{aligned} I & =\int_\limits0^1 \frac{t^2}{\left(1+t^2\right)\left(1+t+t^2\right)} d t \\ & =\int_\limits0^1\left(\frac{x}{1+x^2}-\frac{x}{1+x+x^2}\right) d x \end{aligned}$$

$${1 \over 2}\int\limits_0^1 {{{2x} \over {1 + {x^2}}}dx - \int\limits_0^{} {{{{1 \over 2}(2x + 1) - {1 \over 2}} \over {1 + x + {x^2}}}dx} } $$

$$\begin{aligned} & =\frac{1}{2} \ln 2-\frac{1}{2} \ln 3 \frac{1}{2} \int_\limits0^1 \frac{d x}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\left[\tan ^{-1} \frac{2 x+1}{\sqrt{3}}\right]_0^1 \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \\ & =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} \\ & \therefore \quad a=2, b=6 \\ & \therefore \quad a+b=8 \end{aligned}$$

19:T4ec,

$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} d x=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{\sin ^2 x+\cos ^2 x+\sin x \cos x} d x$$

$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\tan ^2 x}{1+\tan x+\tan ^2 x} d x$$

$$=\int_\limits0^{\frac{\pi}{4}} \frac{\tan x \cdot \sec ^2 x d x}{\left(1+\tan ^2 x\right)\left(1+\tan x+\tan ^2 x\right)}$$

माना $$\tan x=t$$

$$\begin{aligned} I & =\int_\limits0^1 \frac{t^2}{\left(1+t^2\right)\left(1+t+t^2\right)} d t \\ & =\int_\limits0^1\left(\frac{x}{1+x^2}-\frac{x}{1+x+x^2}\right) d x \end{aligned}$$

$${1 \over 2}\int\limits_0^1 {{{2x} \over {1 + {x^2}}}dx - \int\limits_0^{} {{{{1 \over 2}(2x + 1) - {1 \over 2}} \over {1 + x + {x^2}}}dx} } $$

$$\begin{aligned} & =\frac{1}{2} \ln 2-\frac{1}{2} \ln 3 \frac{1}{2} \int_\limits0^1 \frac{d x}{\left(x+\frac{1}{2}\right)^2+\fra

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