_ ^ x^3 1 - x^8 dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}dx = } $

A
$\frac{1}{2}{\sin ^{ - 1}}({x^4}) + c$
B
$\frac{1}{3}{\sin ^{ - 1}}({x^4}) + c$
✓
$\frac{1}{4}{\sin ^{ - 1}}({x^4}) + c$
D
None of these

✓

Answer

Correct option: C.

$\frac{1}{4}{\sin ^{ - 1}}({x^4}) + c$

c
(c)$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}\,dx = \int_{}^{} {\frac{{{x^3}}}{{\sqrt {1 - {{({x^4})}^2}} }}} } $
Put ${x^4} = t \Rightarrow 4{x^3}dx = dt,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{\sqrt {1 - {x^2}} }} = \frac{1}{4}[{{\sin }^{ - 1}}(t)] + c = \frac{1}{4}{{\sin }^{ - 1}}({x^4}) + c.} $

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