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

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