Question

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

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

Answer 1

Answer: (C)

Given, $I=\underset{0}{\overset{\sqrt{\ln (\pi /2)}}{\int }}\cos \left(e^{x^2}\right)\cdot 2x{e}^{x^2}dx$
Put $e^{x^2}=t\Rightarrow 2x{e}^{x^2}dx=dt$
Now, lower limit, $t=1$
Upper limit, $t=e^{\ln \pi /2}=\frac{\pi }{2}$
$\therefore I=\underset{1}{\overset{\pi /2}{\int }}\cos (t)dt=[\sin t{]}_1^{\pi /2}$
$=\sin \left(\frac{\pi }{2}\right)-\sin 1$
$\Rightarrow I=1-\sin 1$