Question

$\displaystyle \int^{\sqrt{\pi}/2}_0$ $2x^{3} sin\left(x^{2}\right) dx =$dx is equal to

• A. $\frac{1}{\sqrt{2}} \left(1+\frac{\pi}{4}\right)$
• B. $\frac{1}{\sqrt{2}} \left(1-\frac{\pi}{4}\right)$
• C. $\frac{1}{\sqrt{2}} \left(\frac{\pi}{2}-1\right)$
• D. $\frac{1}{\sqrt{2}} \left(1-\frac{\pi}{2}\right)$

Answer 1

The Correct Option is B

$ I =\int_{0}^{\sqrt{\pi} / 2} 2 x^{3} \sin \left(x^{2}\right) d x $
$=\int_{0}^{\sqrt{\pi} / 2} 2 x^{2} \cdot x \sin \left(x^{2}\right) d x $
Put $ x^{2}=t $
$\Rightarrow 2 x d x=d t $
Also, when $x=0, $ then $ t=0 $
and when $x=\frac{\sqrt{\pi}}{2}$, then $t=\frac{\pi}{4}$
$\Rightarrow I=\int_{0}^{\pi / 4} t \underset{I}{\text{sin}} \underset{ II }{t} d t$
$=[t(-\cos t)]_{0}^{\pi / 4}-\int_{0}^{\pi / 4}-\cos t(1) d t$
$=[-t \cos t]_{0}^{\pi / 4}+\int_{0}^{\pi / 4} \cos t d t$
$=[-t \cos t+\sin t]_{0}^{\pi / 4}$
$=\left[-\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$
$=\frac{1}{\sqrt{2}}\left(1-\frac{\pi}{4}\right)$