Question

${\int }_0^1\frac{x^7}{\sqrt{1-x^4}}dx$ is equal to

• A. 1
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{1}{3}$

$I={\int }_0^1\frac{x^7}{\sqrt{1-x^4}}dx={\int }_0^1\frac{x^{6}xdx}{\sqrt{1-x^4}}$
$Put{x}^2=sin\theta \Rightarrow 2xdx=cos\theta d\theta$
$I=\frac{1}{2}{}_0^{\frac{\pi }{2}}\frac{si{n}^3\theta .cos\theta d\theta }{cos\theta }=\frac{1}{2}{\int }_0^{\frac{1}{2}}si{n}^3\theta d\theta$
$=\frac{1}{2}\frac{T2T\left(\frac{1}{2}\right)}{2.T\left(\frac{1}{2}\right)}=\frac{T\left(\frac{1}{2}\right)}{4.\frac{3}{2}.\frac{1}{2}.T\left(\frac{1}{2}\right)}=\frac{1}{3}$