Suppose I1=∫0π/2 cos⁡πsin2⁡xdx;I2=∫0π/2 cos⁡2πsin2⁡xdx and I3=∫0π/2 cos⁡(πsin⁡x)dx then,

I1=∫0π/2 cos⁡πsin2⁡xdx∫ab f(x)dx=∫ab f(a+b−x)dxI1=∫0π/2 cos⁡πcos2⁡xdx On adding2I1=∫0π/2 cos⁡πsin2⁡x+cos⁡πcos2⁡xdx=∫0π/2 2cos⁡π2⋅cos⁡π2cos⁡2xdx=0I1=0.......................(i)I2=∫0π/2 cos⁡π(1−cos⁡2x)dx=−∫0π/2 cos⁡(πcos⁡2x)dx=−12∫0π cos⁡(πcos⁡t)dt [Put 2x=t ] =−22∫0π/2 cos⁡(πcos⁡t)dt=−I3∴I2+I3=0.......................(ii) Hence, I1+I2+I3=0

Suppose I1=∫0π/2 cos⁡πsin2⁡xdx;I2=∫0π/2 cos⁡2πsin2⁡xdx and I3=∫0π/2 cos⁡(πsin⁡x)dx then,

I1=∫0π/2 cos⁡πsin2⁡xdx∫ab f(x)dx=∫ab f(a+b−x)dxI1=∫0π/2 cos⁡πcos2⁡xdx On adding2I1=∫0π/2 cos⁡πsin2⁡x+cos⁡πcos2⁡xdx=∫0π/2 2cos⁡π2⋅cos⁡π2cos⁡2xdx=0I1=0.......................(i)I2=∫0π/2 cos⁡π(1−cos⁡2x)dx=−∫0π/2 cos⁡(πcos⁡2x)dx=−12∫0π cos⁡(πcos⁡t)dt [Put 2x=t ] =−22∫0π/2 cos⁡(πcos⁡t)dt=−I3∴I2+I3=0.......................(ii) Hence, I1+I2+I3=0

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Suppose $I_{1} = \int_{0}^{π / 2} \cos ({πsin}^{2} x) dx; I_{2} = \int_{0}^{π / 2} \cos (2 {πsin}^{2} x) dx and I_{3} = \int_{0}^{π / 2} \cos (πsin x) dx$ then,

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$I_{1} = 0$

$I_{2} + I_{3} = 0$

$I_{1} + I_{2} + I_{3} = 0$

$I_{2} = I_{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} I_{1} = \int_{0}^{π / 2} \cos ({πsin}^{2} x) dx \\ \int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx \\ I_{1} = \int_{0}^{π / 2} \cos ({πcos}^{2} x) dx \\ On adding \\ 2 I_{1} = \int_{0}^{π / 2} \cos ({πsin}^{2} x) + \cos ({πcos}^{2} x) dx \\ = \int_{0}^{π / 2} 2 \cos (\frac{π}{2}) \cdot \cos (\frac{π}{2} \cos 2 x) dx = 0 \end{array}$
$\begin{array}{l} I_{1} = 0 . . . . . . . . . . . . . . . . . . . . . . . (i) \\ I_{2} = \int_{0}^{π / 2} \cos (π (1 - \cos 2 x) dx = - \int_{0}^{π / 2} \cos (πcos 2 x) dx) \\ = - \frac{1}{2} \int_{0}^{π} \cos (πcos t) dt [Put 2 x = t] \\ = - \frac{2}{2} \int_{0}^{π / 2} \cos (πcos t) dt = - I_{3} \\ ∴ I_{2} + I_{3} = 0 . . . . . . . . . . . . . . . . . . . . . . . (ii) \\ Hence, I_{1} + I_{2} + I_{3} = 0 \end{array}$