_0^a x (2ax - x^2 ) ^ 3 2 ,dx = - Vidyadip

MCQ

$\int_0^a {x{{(2ax - {x^2})}^{\frac{3}{2}}}\,dx = } $

A
${a^5}\left[ {\frac{{3\pi }}{{16}} - 1} \right]$
B
${a^5}\left[ {\frac{{3\pi }}{{16}} + 1} \right]$
✓
${a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$
D
None of these

✓

Answer

Correct option: C.

${a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$

c
(c) Put $x = a(1 - \cos 2\theta ) $

$\Rightarrow dx = 2a\sin 2\theta \,d\theta $

Therefore, $\int_0^a {x{{(2ax - {x^2})}^{3/2}}dx} $

$ = \int_0^{\pi /4} {2{a^5}(1 - \cos 2\theta ){{\sin }^4}2\theta \,\,d\theta } $

Now again, put $2\theta = \phi $

$ = {a^5}\left[ {\int_0^{\pi /2} {{{\sin }^4}\phi \,d\phi } - \int_0^{\pi /2} {{{\sin }^4}\phi \cos \phi \,d\phi } } \right]$

$ = {a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$.

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