Evaluate: $\int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} dx$

Solution

$I = \int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} dx . . . . . . . . . . . . (i) Using the property \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)}{\sin^{4} (\frac{π}{2} - x) + \cos^{4} (\frac{π}{2} - x)} dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \cos x \sin x}{\sin^{4} x + \cos^{4} x} dx . . . . . . . . . . . . . (ii)$

Adding ( i ) and ( ii ),

$2 I = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} . \sin x \cos x)}{\sin^{4} x + \cos^{4} x} dx \Rightarrow I = \frac{π}{4} \int_{0}^{\frac{π}{2}} [\frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x}] dx = \frac{π}{4} \int_{0}^{\frac{π}{2}} [\frac{\frac{\sin x \cos x}{\cos^{4} x}}{\frac{\sin^{4} x}{\cos^{4} x} + 1}] dx = \frac{π}{4} \int_{0}^{\frac{π}{2}} \frac{\tan x \sec^{2} x}{\tan^{4} x + 1} dx$

Put tan² x = z

$∴$ 2 tan x sec² x dx = dz

$\Rightarrow \tan x \sec^{2} x dx = \frac{dz}{2} When x = 0, z = 0 and when x = \frac{π}{2}, z = \infty ∴ I = \frac{π}{4} \int_{0}^{\infty} \frac{\frac{dz}{2}}{z^{2} + 1} \Rightarrow I = \frac{π}{8} \int_{0}^{\infty} \frac{dz}{1 + z^{2}} = \frac{π}{8} {|\tan^{- 1} (z)|}_{0}^{\infty} = \frac{π}{8} \tan^{- 1} \infty - \tan^{- 1} 0 = \frac{π}{8} (\frac{π}{2} - 0) = \frac{π^{2}}{16}$

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Integrals

Evaluate: $\int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} dx$

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Evaluate: $\int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} dx$