If f(x)=sin ^6 x+cos ^6 x+2 sin ^3 x cos ^3 x | vedclass

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Question

(B) Given, $f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x = (\sin ^3 x+\cos ^3 x)^2$. We need to evaluate $I=\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Given, $f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x = (\sin ^3 x+\cos ^3 x)^2$. We need to evaluate $I=\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x$. Using $\sin 2x = 2 \sin x \cos x$, we have $\sin ^2 2x = 4 \sin ^2 x \cos ^2 x$. Thus, $I = \int_0^{\pi / 4} \frac{4 \sin ^2 x \cos ^2 x}{(\sin ^3 x+\cos ^3 x)^2} d x$. Dividing numerator and denominator by $\cos ^6 x$, we get: $I = \int_0^{\pi / 4} \frac{4 	an ^2 x \sec ^2 x}{(	an ^3 x+1)^2} d x$. Let $t = 	an ^3 x$, then $dt = 3 	an ^2 x \sec ^2 x dx$, which implies $	an ^2 x \sec ^2 x dx = \frac{dt}{3}$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$. Substituting these into the integral: $I = \int_0^1 \frac{4 \cdot (dt/3)}{(t+1)^2} = \frac{4}{3} \int_0^1 (t+1)^{-2} dt$. $I = \frac{4}{3} \left[ \frac{-1}{t+1} ight]_0^1 = \frac{4}{3} \left( -\frac{1}{2} - (-1) ight) = \frac{4}{3} \left( \frac{1}{2} ight) = \frac{2}{3}$.

If $f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x$, then $\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x=$

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