MDoubt No 000004 - JEE

Question : If $\displaystyle\int\limits_{0}^{1}\frac{dx}{1+x^{3/2}}$, then

(a) $\ln 2 < I < \dfrac{\pi}{4}$
(b) $\log_{10} 2 < I < \ln 2$
(c) $\dfrac{\pi}{4} <I, \ln 2 < I$
(d) $\dfrac{\pi}{4} >I, \ln 2> I$

Solution : Given $I=\displaystyle\int\dfrac{x^3\sin^{-1}x^2}{\sqrt{1-x^4}}dx$
Let $\sin^{-1}x^2=t\implies x^2=\sin t$
$\implies d\sin^{-1}x^2=dt$
$\implies\dfrac{1}{\sqrt{1-x^4}}\cdot 2xdx=dt$
$\therefore I=\displaystyle\int\frac{t\sin t}{2}dt$
$\implies I=\dfrac{\sin t-t\cos t}{2}+C$
$\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\cdot\cos\left(\sin^{-1}x^2\right)}{2}+C$
$\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\cdot\cos\left(\cos^{-1}\sqrt{1-x^4}\right)}{2}+C$
$\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\sqrt{1-x^4}}{2}+C$