Suppose I1=∫0π/2cos(π sin2x)dx; I2=∫0π/2cos(2π sin 2x)dx and I3=∫0π/2cos(π sin x)dx, then

We have ,I1=∫0π/2cos(π sin2dx)dxI1=∫0π/2cos(π cos2x)dxOn adding 2I1=∫0π/2cos (π sin2x)+cos (π cos2x)dx=∫0π/22 cos (π2).cos(π2cos 2 x)dx=0⇒ I1=0I2=∫0π/2cos{π(1−cos 2x)}dx=−∫0π/2cos(π cos 2x)dx=−12∫0π/2cos (π cos t)dt=I3⇒I2+I3=0I3=−∫0π/2cos (π sin t)dt∴I2+I3=0Hence, I1+I2+I3=0

Suppose I1=∫0π/2cos(π sin2x)dx; I2=∫0π/2cos(2π sin 2x)dx and I3=∫0π/2cos(π sin x)dx, then

We have ,I1=∫0π/2cos(π sin2dx)dxI1=∫0π/2cos(π cos2x)dxOn adding 2I1=∫0π/2cos (π sin2x)+cos (π cos2x)dx=∫0π/22 cos (π2).cos(π2cos 2 x)dx=0⇒ I1=0I2=∫0π/2cos{π(1−cos 2x)}dx=−∫0π/2cos(π cos 2x)dx=−12∫0π/2cos (π cos t)dt=I3⇒I2+I3=0I3=−∫0π/2cos (π sin t)dt∴I2+I3=0Hence, I1+I2+I3=0

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Suppose $I_{1} = \int_{0}^{π / 2} \cos (π \sin^{2} x) d x; I_{2} = \int_{0}^{π / 2} \cos (2 π \sin^{2} x) d x a n d I_{3} = \int_{0}^{π / 2} \cos (π \sin x) d x, t h e n$

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$I_{1} \neq I_{2} + I_{3}$

$I_{1} < I_{2} + I_{3}$

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

$I_{1} > I_{2} + I_{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} W e h a v e, I_{1} = \int_{0}^{π / 2} \cos (π \sin^{2} d x) d x \\ I_{1} = \int_{0}^{π / 2} \cos (π \cos^{2} x) d x \\ O n a d d i n g 2 I_{1} = \int_{0}^{π / 2} \cos (π \sin^{2} x) + c o s (π \cos^{2} x) d x \\ = \int_{0}^{π / 2} 2 \cos (\frac{π}{2}) . \cos (\frac{π}{2} \cos 2 x) d x = 0 \\ \Rightarrow I_{1} = 0 \\ I_{2} = \int_{0}^{π / 2} \cos {π (1 - \cos 2 x)} d x \\ = - \int_{0}^{π / 2} \cos (π \cos 2 x) d x \\ = - \frac{1}{2} \int_{0}^{π / 2} \cos (π \cos t) d t = I_{3} \\ \Rightarrow I_{2} + I_{3} = 0 \\ I_{3} = - \int_{0}^{π / 2} \cos (π \sin t) d t \\ ∴ I_{2} + I_{3} = 0 \\ H e n c e, I_{1} + I_{2} + I_{3} = 0 \end{array}$