Question

Suppose $I_1={\int }_0^{\frac{\pi }{2}}cos\left(\pi si{n}^{2}x\right)dx,I_2={\int }_0^{\frac{\pi }{2}}cos\left(2\pi si{n}^{2}x\right)dxand{I}_3={\int }_0^{\frac{\pi }{2}}cos\left(\pi sinx\right)dx,$ then

• A. $I_1=0$
• B. $I_2+I_3=0$
• C. $I_1+I_2+I_3=0$
• D. $I_1+I_2+I_3=1$
  

Answer 1

Answer: (A), (B)

The correct options are
A $I_1=0$
B $I_2+I_3=0$

$I_1={\int }_0^{\frac{\pi }{2}}cos\left(\pi si{n}^{2}x\right)dx$
apply 'a + b - x' property
$I_1={\int }_0^{\frac{\pi }{2}}cos\left(\pi co{s}^{2}x\right)dx$
On adding
$2I_1={\int }_0^{\frac{\pi }{2}}2cos\left(\frac{\pi }{2}\right).cos\left(\frac{\pi }{2}cos2x\right)dx=0$
$\Rightarrow I_1=0$ ...(i)
$I_2={\int }_0^{\frac{\pi }{2}}cos\left(\pi \right(1-cos2x)dx=-{\int }_0^{\frac{\pi }{2}}cos(\pi cos2x)dx$
$=-\frac{1}{2}{\int }_0^{\pi }cos\left(\pi cost\right)dt$ [Put 2x = t]
$=-\frac{2}{2}{\int }_0^{\frac{\pi }{2}}cos\left(\pi cost\right)dt=-I_3\Rightarrow I_2+I_3=0$ ...(ii)