Question

The value of $\int\limits_{0}^{\sqrt{\ln (\pi / 2)}} \cos \left(e^{x^{2}}\right) 2 x e^{x^{2}} d x$ is

• A. $1$
• B. $1+\sin 1$
• C. $1-\sin 1$
• D. $(\sin 1)-1$

Answer 1

Answer: (C)

Given, $I=\int\limits_{0}^{\sqrt{\ln (\pi / 2)}} \cos \left(e^{x^{2}}\right) \cdot 2 x e^{x^{2}} d x$
Put $e^{x^{2}}=t \Rightarrow 2 x e^{x^{2}} d x=d t$
Now, lower limit, $t=1$
Upper limit, $t=e^{\ln \pi / 2}=\frac{\pi}{2}$
$\therefore I=\int\limits_{1}^{\pi / 2} \cos (t) d t=[\sin t]_{1}^{\pi / 2}$
$=\sin \left(\frac{\pi}{2}\right)-\sin 1$
$\Rightarrow I=1-\sin 1$