_ ^ ^ - 1 (3x - 4 x^3 )dx = - Vidyadip

MCQ

$\int_{}^{} {{{\sin }^{ - 1}}(3x - 4{x^3})dx = } $

A
$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + c$
B
$x{\sin ^{ - 1}}x - \sqrt {1 - {x^2}} + c$
C
$2[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$
✓
$3[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$

✓

Answer

Correct option: D.

$3[x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} ] + c$

d
(d) Put $x = \sin \theta \Rightarrow dx = \cos \theta \,d\theta ,$ therefore
$\int_{}^{} {{{\sin }^{ - 1}}(3x - 4{x^3})} \,dx = \int_{}^{} {{{\sin }^{ - 1}}(\sin 3\theta )\cos \theta \,d\theta } $
$ = \int_{}^{} {3\theta \cos \theta \,d\theta } = 3\left\{ {\theta \sin \theta - \int_{}^{} {\sin \theta \,d\theta } } \right\}$
$ = 3\left\{ {\theta \sin \theta + \cos \theta } \right\} + c = 3\left\{ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right\} + c.$

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