Q101 / 170 · Manipal (MET) 2020 · Mathematics
The value of $\int_0^{\sqrt{\ln (\pi / 2)}} \cos \left(e^{x^2}\right) 2 x e^{x^2} d x$ is
- A$1$
- B$1+\sin 1$
- Correct answerC$1-\sin 1$
- D$(\sin 1)-1$
Solution
Given, $I=\int_0^{\sqrt{\ln (\pi / 2)}} \cos \left(e^{x^2}\right) \cdot 2 x e^{x^2} d x$
Put $\quad e^{x^2}=t \Rightarrow 2 x e^{x^2} d x=d t$
Now, lower limit, $t=1$
Upper limit, $t=e^{\ln \pi / 2}=\frac{\pi}{2}$
$\therefore \quad I=\int_1^{\pi / 2} \cos (t) d t=[\sin t]_1^{\pi / 2}$
$=\sin \left(\frac{\pi}{2}\right)-\sin 1$
$\Rightarrow \quad I=1-\sin 1$
Hence, the correct answer is option: 'C'.