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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


ANSWER

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


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


ANSWER

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


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
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