| 120 3 ? 0 x 2 sin x cos x sin 4 x + cos 4 x ? d x | is equal to [2024]

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Solution

Q.
$| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ is equal to ________ . [2024]

Ans.
(15)

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

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

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

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

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

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

$= \frac{- π^{2}}{2} \int_{0}^{π / 2} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$

$Put \cos 2 x = t ∴ I = \frac{- π^{2}}{2} \int_{1}^{- 1} \frac{- d t}{2 (1 + t^{2})}$

$= \frac{- π^{2}}{2} \int_{0}^{1} \frac{d t}{1 + t^{2}} = \frac{- π^{2}}{2} {(\tan^{- 1} t)}_{0}^{1} = \frac{- π^{2}}{2} (\frac{π}{4}) = \frac{- π^{3}}{8}$

$∴$ $| \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} |$ $= \frac{120}{π^{3}} \times \frac{π^{3}}{8} = 15$

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