If 0 4 sin 2 x 1 + sin x cos x ? d x = 1 a log e ( a 3 ) + b 3 , where a , b N , then a + b is equal to ______ . [2024]

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Solution

Q.
If $\int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x = \frac{1}{a} \log_{e} (\frac{a}{3}) + \frac{π}{b \sqrt{3}}, where a, b \in N,$ then $a + b$ is equal to ______ . [2024]

Ans.
(8)

Let $I = \int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{1 + \sin x \cos x} d x$

$= \int_{0}^{\frac{π}{4}} \frac{\sin^{2} x}{\sin^{2} x + \cos^{2} x + \sin x \cos x} d x$

$\Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\tan^{2} x}{1 + \tan x + \tan^{2} x} d x$

$= \int_{0}^{\frac{π}{4}} \frac{\tan^{2} x \cdot \sec^{2} x d x}{(1 + \tan^{2} x) (1 + \tan x + \tan^{2} x)}$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$I = \int_{0}^{1} \frac{t^{2}}{(1 + t^{2}) (1 + t + t^{2})} d t = \int_{0}^{1} (\frac{t}{1 + t^{2}} - \frac{t}{1 + t + t^{2}}) d t$

$= \frac{1}{2} \int_{0}^{1} \frac{2 t}{1 + t^{2}} d t - \int_{0}^{1} \frac{\frac{1}{2} (2 t + 1) - \frac{1}{2}}{1 + t + t^{2}} d t$

$= \frac{1}{2} \ln 2 - \frac{1}{2} \ln 3 + \frac{1}{2} \int_{0}^{1} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$

$= \frac{1}{2} \ln \frac{2}{3} + \frac{1}{2} \cdot \frac{2}{\sqrt{3}} {[\tan^{- 1} \frac{2 x + 1}{\sqrt{3}}]}_{0}^{1} = \frac{1}{2} \ln \frac{2}{3} + \frac{1}{\sqrt{3}} (\frac{π}{3} - \frac{π}{6})$

$= \frac{1}{2} \ln \frac{2}{3} + \frac{1}{\sqrt{3}} \cdot \frac{π}{6}$ $∴ a = 2, b = 6$

$∴ a + b = 8$

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