The value of the integral 0 / 4 x ? d x sin 4 ( 2 x ) + cos 4 ( 2 x ) equals: [2024]

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Solution

Q.
$The value of the integral \int_{0}^{π / 4} \frac{x d x}{\sin^{4} (2 x) + \cos^{4} (2 x)} equals:$ [2024]

1 $\frac{\sqrt{2} π^{2}}{16}$

2 $\frac{\sqrt{2} π^{2}}{8}$

3 $\frac{\sqrt{2} π^{2}}{64}$

4 $\frac{\sqrt{2} π^{2}}{32}$

Ans.
(4)

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

Put $2 x = t$

$\Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2}$ $∴ I = \int_{0}^{π / 2} \frac{t / 2 \cdot d t / 2}{\sin^{4} t + \cos^{4} t}$

$\Rightarrow I = \frac{1}{4} \int_{0}^{π / 2} \frac{t d t}{\sin^{4} t + \cos^{4} t}$ ...(i)

$\Rightarrow I = \frac{1}{4} \int_{0}^{π / 2} \frac{(\frac{π}{2} - t) d t}{\sin^{4} t + \cos^{4} t}$ ...(ii)

Adding (i) and (ii), we get

$2 I = \frac{π}{8} \int_{0}^{π / 2} \frac{d t}{\sin^{4} t + \cos^{4} t} \Rightarrow 2 I = \frac{π}{8} \int_{0}^{π / 2} \frac{\sec^{4} t d t}{\tan^{4} t + 1}$

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

$∴ 2 I = \frac{π}{8} \int_{0}^{\infty} \frac{(1 + y^{2}) d y}{1 + y^{4}}$

$\Rightarrow 2 I = \frac{π}{8} \int_{0}^{\infty} \frac{(\frac{1}{y^{2}} + 1) d y}{y^{2} + \frac{1}{y^{2}} - 2 + 2} = \frac{π}{8} \int_{0}^{\infty} \frac{(\frac{1}{y^{2}} + 1) d y}{{(y - \frac{1}{y})}^{2} + 2}$

Let $y - \frac{1}{y} = u \Rightarrow (1 + \frac{1}{y^{2}}) d y = d u,$ we get $2 I = \frac{π}{8} \int_{- \infty}^{\infty} \frac{d u}{u^{2} + 2}$

$\Rightarrow 2 I = \frac{π}{8} {[\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}})]}_{- \infty}^{\infty}$

$\Rightarrow 2 I = \frac{π}{8} [\frac{1}{\sqrt{2}} \tan^{- 1} (\infty) - \frac{1}{\sqrt{2}} \tan^{- 1} (- \infty)]$

$\Rightarrow I = \frac{π}{16 \sqrt{2}} (\frac{π}{2} + \frac{π}{2}) = \frac{π^{2}}{16 \sqrt{2}} = \frac{\sqrt{2} π^{2}}{32}$

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