If I = 0 2 sin 3 2 x sin 3 2 x + cos 3 2 x d x , then 0 2 I x sin x cos x sin 4 x + cos 4 x d x equals: [2025]

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Solution

Q.
If $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ , then $\int_{0}^{2 I} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ equals: [2025]

1 $\frac{π^{2}}{12}$

2 $\frac{π^{2}}{16}$

3 $\frac{π^{2}}{8}$

4 $\frac{π^{2}}{4}$

Ans.
(2)

We have, $I = \int_{0}^{π / 2} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ ... (i)

$I = \int_{0}^{π / 2} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} d x$ ... (ii)

Adding equation (i) and (ii), we get

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

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

$\Rightarrow I_{2} = \int_{0}^{π / 2} \frac{(\frac{π}{2} - x) \sin x \cos x d x}{\cos^{4} x + \sin^{4} x}$ (iv)

Adding (iii) and (iv), we get

$I_{2} = \frac{π}{4} \int_{0}^{π / 2} \frac{\tan x \sec^{2} x d x}{\tan^{4} x + 1}$

Now, put $\tan^{2} x = t$

$I_{2} = \frac{π}{8} \int_{0}^{\infty} \frac{d t}{t^{2} + 1} = \frac{π}{8} \tan^{- 1} (t) |_{0}^{\infty} = \frac{π}{8} \cdot \frac{π}{2} = \frac{π^{2}}{16}$

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