#### Passage

Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$

On the basis of above information, answer the following questions:
Mathematics

# $I_{2}$ is equal to

$\displaystyle \frac {3}{2}\int_{0}^{\pi/2}\frac {\sin^{3}\theta \cos\theta}{\sqrt{1-\sin^{3}\theta}}d\theta$

##### SOLUTION
$\displaystyle I_{2}\int_{0}^{1}(1-x^{3})^{1/2}dx$

$\displaystyle =x(1-x^{3})^{1/2})^{1}_{0}+\frac {3}{2}\int_{0}^{1}\frac {x^{3}dx}{\sqrt{1-x^{3}}}$

Put $x=sin\theta$

$\displaystyle I_{2}=\frac {3}{2}\int_{0}^{\pi/2}\frac {sin^{3}\theta cos\theta}{\sqrt{1-sin^{3}\theta}}d\theta$

Ans: $C$

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Single Correct Medium Published on 17th 09, 2020
Mathematics

# $I_{1}$ is equal to

$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$

##### SOLUTION
$\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx$

$x=sin\theta, dx=cos\theta d\theta$

$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta$

$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta$

Ans: $A$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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