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Question: \[\int_{0}^{a}{x(2ax - x^{2})^{\frac{3}{2}}dx =}\]...

∫0ax(2ax−x2)32dx=\int_{0}^{a}{x(2ax - x^{2})^{\frac{3}{2}}dx =}

A

a5[3π16−1]a^{5}\left\lbrack \frac{3\pi}{16} - 1 \right\rbrack

B

a5[3π16+1]a^{5}\left\lbrack \frac{3\pi}{16} + 1 \right\rbrack

C

a5[3π16−15]a^{5}\left\lbrack \frac{3\pi}{16} - \frac{1}{5} \right\rbrack

D

None of these

Answer

a5[3π16−15]a^{5}\left\lbrack \frac{3\pi}{16} - \frac{1}{5} \right\rbrack

Explanation

Solution

Put x=a(1−cos⁡2θ)⇒dx=2asin⁡2θdθx = a(1 - \cos 2\theta) \Rightarrow dx = 2a\sin 2\theta d\theta

Therefore, ∫0ax(2ax−x2)3/2dx\int_{0}^{a}{x(2ax - x^{2})^{3/2}dx}

=∫0π/42a5(1−cos⁡2θ)sin⁡42θdθ= \int_{0}^{\pi/4}{2a^{5}(1 - \cos 2\theta)\sin^{4}2\theta d\theta}

Now again, put 2θ=φ2\theta = \varphi

=a5[∫0π/2sin⁡4φdφ−∫0π/2sin⁡4φcos⁡φdφ]=a5[3π16−15]= a^{5}\left\lbrack \int_{0}^{\pi/2}{\sin^{4}\varphi d\varphi} - \int_{0}^{\pi/2}{\sin^{4}\varphi\cos\varphi d\varphi} \right\rbrack = a^{5}\left\lbrack \frac{3\pi}{16} - \frac{1}{5} \right\rbrack.