_0^a x^4 ,dx ( a^2 + x^2 ) ^4 = - Vidyadip

MCQ

$\int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} = $

✓
$\frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$
B
$\frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} + \frac{1}{3}} \right)$
C
$\frac{1}{{16}}{a^3}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$
D
$\frac{1}{{16}}{a^3}\left( {\frac{\pi }{4} + \frac{1}{3}} \right)$

✓

Answer

Correct option: A.

$\frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$

a
(a) Put $x = a\tan \theta $

$\Rightarrow dx = a{\sec ^2}\theta \,d\theta ,$ then we have

$I = \int_0^{\pi /4} {\frac{{{a^4}{{\tan }^4}\theta .\,a{{\sec }^2}\theta \,d\theta }}{{{a^8}{{\sec }^8}\theta }}} $

==> $\frac{1}{{{a^3}}}\int_0^{\pi /4} {{{\sin }^4}\theta } {\cos ^2}\theta \,d\theta $

$= I = \frac{1}{{{a^3}}}\left[ {\int_0^{\pi /4} {({{\sin }^4}\theta } - {{\sin }^6}\theta } \right]\,d\theta $

$ = \frac{1}{{{a^3}}}\int_0^{\pi /4} {\left[ {\frac{{{{(1 - \cos 2\theta )}^2}}}{4} - \frac{{{{(1 - \cos 2\theta )}^3}}}{8}} \right]\,} d\theta $

$ = \frac{1}{{8{a^3}}}\int_0^{\pi /4} {{\rm{ }}(1 + \cos 2\theta } )(1 + {\cos ^2}2\theta - 2\cos 2\theta )d\theta $

$ = \frac{1}{{8{a^3}}}\int_0^{\pi /4} {(1 - \cos 2\theta - {{\cos }^2}2\theta + {{\cos }^3}2\theta )\,d\theta } $

$ = \frac{1}{{32{a^3}}}\int_0^{\pi /4} {(2 - \cos 2\theta - 2\cos 4\theta + \cos 6\theta )d\theta } $

$ = \frac{1}{{32{a^3}}}\left[ {2\theta - \frac{{\sin 2\theta }}{2} - \frac{{\sin 4\theta }}{2} + \frac{{\sin 6\theta }}{6}} \right]_0^{\pi /4}$

$ = \frac{1}{{16{a^3}}}\left( {\frac{\pi }{4} - \frac{1}{3}} \right)$.

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