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Mathematics Question on Integrals of Some Particular Functions

The value of the integral ∫0π/2sin⁡5 x dx\int\limits^{\pi/2}_{0}\sin^{5}\, x\,dx is

A

415\frac{4}{15}

B

85\frac{8}{5}

C

815\frac{8}{15}

D

45\frac{4}{5}

Answer

815\frac{8}{15}

Explanation

Solution

I=∫0π/2sin⁡4x⋅sin⁡x dxI=\int\limits_{0}^{\pi / 2} \sin ^{4} x \cdot \sin x \,d x
=∫0π/2(1−cos⁡2x)2sin⁡x dx=\int\limits_{0}^{\pi / 2}\left(1-\cos ^{2} x\right)^{2} \sin x\, d x
Put cos⁡x=t\cos x=t
⇒−sin⁡x dx=dt\Rightarrow -\sin x\, d x=d t
=−∫10(1−t2)2dt=∫01(t4−2t2+1)dt=-\int\limits_{1}^{0}\left(1-t^{2}\right)^{2} d t=\int\limits_{0}^{1}\left(t^{4}-2 t^{2}+1\right) d t
[∵−∫abf(x)dx=∫abf(x)dx]\left[\because-\int_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(x) d x\right]
=15(t5)01−23(t3)01+(t)01=\frac{1}{5}\left(t^{5}\right)_{0}^{1}-\frac{2}{3}\left(t^{3}\right)_{0}^{1}+(t)_{0}^{1}
=15−23+1=3−10+1515=815=\frac{1}{5}-\frac{2}{3}+1=\frac{3-10+15}{15}=\frac{8}{15}