Question

The value of $\underset{0}{\overset{\pi /2}{\int }}\left(\cos \right(2018\pi {\sin }^{2}x)+\cos (1009\pi \sin x\left)\right)dx$ is

• A. $0$
• B. $1$
• C. $\frac{2}{\pi }$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $0$

$\underset{0}{\overset{\pi /2}{\int }}\left[\cos \right(2018\pi {\sin }^{2}x)+\cos (1009\pi \sin x\left)\right]dx$
Let $I_1=\underset{0}{\overset{\pi /2}{\int }}\cos \left(2018\pi {\sin }^{2}x\right)dx$
$I_2=\underset{0}{\overset{\pi /2}{\int }}\cos \left(1009\pi \sin x\right)dx$
Now,
$I_1=\underset{0}{\overset{\pi /2}{\int }}\cos \left(1009\pi \right(1-\cos 2x\left)\right)dx\Rightarrow I_1=-\underset{0}{\overset{\pi /2}{\int }}\cos \left(1009\pi \cos 2x\right)dx\Rightarrow I_1=-2\underset{0}{\overset{\pi /4}{\int }}\cos \left(1009\pi \cos 2x\right)dx$
Assuming $2x=t\Rightarrow 2dx=dt$
$\Rightarrow I_1=-\underset{0}{\overset{\pi /2}{\int }}\cos \left(1009\pi \cos t\right)dt$
Putting $x\to \frac{\pi }{2}-x$, we get
$\Rightarrow I_1=-I_2\therefore I_1+I_2=0$