Question

Evaluate: $\int \frac{x\sqrt{x}.dx}{\sqrt{1-x^5}}$

• A. $\frac{2}{5}x{\sin }^{-1}\left(x^{\frac{5}{2}}\right)+c$
• B. $\frac{1}{5}x{\sin }^{-1}\left(x^{\frac{3}{2}}\right)+c$
• C. $\frac{-2}{5}{\sin }^{-1}\left(\sqrt{1-x^5}\right)+c$
• D. $\frac{1}{3}x{\sin }^{-1}\left(x^{\frac{3}{2}}\right)+c$

Answer 1

Answer: (C)

$\frac{-2}{5}{\sin }^{-1}\left(\sqrt{1-x^5}\right)+c$

$\frac{x\sqrt{x}}{\sqrt{1-x^5}}dx$

put $1-x^5=t^2,x^5=1-t^2,x=(1-t^2{)}^{1/5}$

$-5x^{4}dx=2tdt$

$dx=\frac{-2t}{5x^4}dt$

$dx=\frac{-2}{5}\frac{t}{(1-t^2{)}^{4/5}}dt$

$\Rightarrow \int \frac{x\sqrt{x}}{\sqrt{1-x^5}}dx=-\int \frac{(1-t^2{)}^{3/10}}{t}\frac{2}{5}\frac{t}{(1-t^2)4/5}dt$

$=\frac{-2}{5}\int \frac{(1-t^2{)}^{3/10}}{(1-t^2{)}^{4/5}}dt$

$=\frac{-2}{5}\int \frac{(1-t^2{)}^{4/5}(1-t^2{)}^{\frac{-1}{2}}}{(1-t^2{)}^{4/5}}dt$

$=\frac{-2}{5}\int \frac{1}{\sqrt{1-t^2}}dt=\frac{-2}{5}{\sin }^{-1}\left(t\right)+c$

$=\frac{-2}{3}{\sin }^{-1}\left(\sqrt{1-x^5}\right)+c$