If / 2 / 2 8 2 cos x ? d x ( 1 + e sin x ) ( 1 + sin 4 x ) = log e ( 3 + 2 2 ) , where are integers, then 2 + 2 equals ____ . [2024]

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Solution

Q.
$If \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} = α π + β \log_{e} (3 + 2 \sqrt{2}),$ $where α, β are integers, then α^{2} + β^{2} equals ____ .$ [2024]

Ans.
(8)

Let $I = \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x d x}{(1 + e^{\sin x}) (1 + \sin^{4} x)}$

Using $\int_{- a}^{a} f (a + b - x) d x = \int_{- a}^{a} f (x) d x$

$\Rightarrow 2 I = \int_{- π / 2}^{π / 2} (\frac{8 \sqrt{2} \cos x}{(1 + e^{\sin x}) (1 + \sin^{4} x)} + \frac{e^{\sin x} 8 \sqrt{2} \cos x}{(e^{\sin x} + 1) (1 + \sin^{4} x)}) d x$

$= \int_{- π / 2}^{π / 2} \frac{8 \sqrt{2} \cos x}{1 + \sin^{4} x} d x = 2 \int_{0}^{π / 2} \frac{8 \sqrt{2} \cos x}{1 + \sin^{4} x} d x$ $[∵ \int_{- a}^{a} f (x) d x = \int_{0}^{a} f (x) d x, if f (- x) = f (x)]$

Put $t = \sin x \Rightarrow d t = \cos x d x$

$∴ I = 8 \sqrt{2} \int_{0}^{1} \frac{d t}{1 + t^{4}} = 4 \sqrt{2} \int_{0}^{1} \frac{2 d t}{1 + t^{4}}$

$= 4 \sqrt{2} \int_{0}^{1} (\frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2}}) d t - 4 \sqrt{2} \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t$

$= 4 \sqrt{2} \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} d t - 4 \sqrt{2} \frac{1}{2 \sqrt{2}} {[\ln \frac{(1 + \frac{1}{t} - \sqrt{2})}{t + \frac{1}{t} + \sqrt{2}}]}_{0}^{1}$

$= 4 \sqrt{2} \cdot \frac{1}{\sqrt{2}} \tan^{- 1} {(\frac{t - \frac{1}{t}}{\sqrt{2}})}_{0}^{1} - \frac{4 \sqrt{2}}{2 \sqrt{2}} {[\ln \frac{(t + \frac{1}{t} - \sqrt{2})}{(t + \frac{1}{t} + \sqrt{2})}]}_{0}^{1}$

$= 2 π + 2 \ln (3 + 2 \sqrt{2}) = α π + β \log_{e} (3 + 2 \sqrt{2})$

$\Rightarrow α = β = 2$

$Then α^{2} + β^{2} = 4 + 4 = 8$

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