If λ=|120π3∫0πx2sinxcosxsin4x+cos4x dx| then λ3=__________

=∫π6π31−sin2xdx =∫π6π3|sinx−cosx|dx =∫π6π4(cosx−sinx)dx+∫π4π3(sinx−cosx)dx =−1+22−3 =α+β2+γ3 =α−1,β=2,γ=−1 α+β+γ=0

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$I f λ = | \frac{120}{π^{3}} \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x |$ then $\frac{λ}{3} =$ __________

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answer is 5.

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Detailed Solution

$= \int_{\frac{π}{6}}^{\frac{π}{3}} \sqrt{1 - \sin 2 x} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} | \sin x - \cos x | d x = \int_{\frac{π}{6}}^{\frac{π}{4}} (\cos x - \sin x) d x + \int_{\frac{π}{4}}^{\frac{π}{3}} (\sin x - \cos x) d x = - 1 + 2 \sqrt{2} - \sqrt{3} = α + β \sqrt{2} + γ \sqrt{3} = α - 1, β = 2, γ = - 1 α + β + γ = 0$