MCQ
$\int_0^1 {\frac{{{x^7}}}{{\sqrt {1 - {x^4}} }}dx} $ is equal to
- A$1$
- ✓$\frac{1}{3}$
- C$\frac{2}{3}$
- D$\frac{\pi }{3}$
✓
Answer
Correct option: B.
$\frac{1}{3}$
b
(b) $I = \int_0^1 {\frac{{{x^7}}}{{\sqrt {1 - {x^4}} }}dx = \int_0^1 {\frac{{{x^6}x\,dx}}{{\sqrt {1 - {x^4}} }}} } $
(b) $I = \int_0^1 {\frac{{{x^7}}}{{\sqrt {1 - {x^4}} }}dx = \int_0^1 {\frac{{{x^6}x\,dx}}{{\sqrt {1 - {x^4}} }}} } $
Put ${x^2} = \sin \theta $ $ \Rightarrow 2x\,dx = \cos \theta \,d\theta $
$I = \frac{1}{2}\int_0^{\pi /2} {\frac{{{{\sin }^3}\theta .\cos \theta \,\,d\theta }}{{\cos \theta }}} $
$= \frac{1}{2}\int_0^{\pi /2} {{{\sin }^3}\theta \,\,d\theta } $
$ = \frac{1}{2}\frac{{\Gamma 2\,\Gamma (1/2)}}{{2.\Gamma (5/2)}} $
$= \frac{{\Gamma \left( {\frac{1}{2}} \right)}}{{4.\frac{3}{2}.\frac{1}{2}.\Gamma \left( {\frac{1}{2}} \right)}} = \frac{1}{3}$.
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