Question

${\int }_0^1\frac{4x^3}{\sqrt{1-x^8}}dx=?$

• A. $\pi$
• B. $-\pi$
• C. $\pi /2$
• D. $-\pi /2$

Answer 1

The correct option is C $\pi /2$

${\int }_0^1\frac{4x^3}{\sqrt{1-x^8}}dx={\int }_0^1\frac{d\left(x^4\right)}{\sqrt{1-(x^4{)}^2}}$

$={\int }_0^1\frac{d\left(x^4\right)}{\sqrt{1-(x^4{)}^2}}=\left[si{n}^{-1}\right(x^4)+c{]}_0^1(\int \frac{1}{\sqrt{1-x^2}}=si{n}^{-1}x)$

$=\left[\right(si{n}^{-1}\left(1\right)+c)-(si{n}^{-1}\left(0\right)+c\left)\right]$

$=si{n}^{-1}1=\frac{\pi }{2}$