I=∫0π/2x sin8 2x dx =0+π/22∫0π/2sin8 2x dx=π4 ∫0π/2sin8 2x dx ∴ ∫0a x f(x) dx=a2 ∫0a f(x) dx if f(a-x)=f(x) here sin8 2π2-x=sin8 2x I=π4 ∫0π sin8 tdt2 put 2x=t ⇒ dx=dt2 =π8 ∫0π sin8 t dt =π8.2 ∫0π/2 sin8 t dt ∵ sin8 (π-t)=sin8t =π478 56 34 12 π2 =35π21024

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$\int_{0}^{π / 2} x \sin^{8} 2 x d x =$

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$\frac{3 π^{2}}{128}$

$\frac{35 π^{2}}{1024}$

$\frac{π^{2}}{32}$

$\frac{5 π^{2}}{32}$

answer is A.

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Detailed Solution

$I = \int_{0}^{π / 2} x \sin^{8} 2 x dx = \frac{0 + π / 2}{2} \int_{0}^{π / 2} \sin^{8} 2 x dx = \frac{π}{4} \int_{0}^{π / 2} \sin^{8} 2 x dx (∴ \int_{0}^{a} x f (x) dx = \frac{a}{2} \int_{0}^{a} f (x) dx if f (a - x) = f (x) here \sin^{8} 2 (\frac{π}{2} - x) = \sin^{8} 2 x) I = \frac{π}{4} \int_{0}^{π} \sin^{8} t (\frac{dt}{2}) put 2 x = t \Rightarrow dx = \frac{dt}{2} = \frac{π}{8} \int_{0}^{π} \sin^{8} t dt = \frac{π}{8} . 2 \int_{0}^{π / 2} \sin^{8} t dt (∵ \sin^{8} (π - t) = \sin^{8} t) = \frac{π}{4} (\frac{7}{8} \frac{5}{6} \frac{3}{4} \frac{1}{2} \frac{π}{2}) = \frac{35 π^{2}}{1024}$