(In reply to
Puzzle Answer by K Sengupta)
We know that:
sin (pi/2 - t) = cos t, and cos(pi/2 - t)= sin(t)
Also,
I (0, pi/2) f(x) dx = I (0, pi/2 - x) f(x) dx
Then, we must have:
I (0, pi/2) {sin^2(sinx) + cos^2(cos x)} dx
= I(0, pi/2) {sin^2(pi/2 - sin x) + cos^2(pi/2 - cos x)} dx
= I(0, pi/2) { cos^2(sin x) + sin^2(cos x)}
= J (say)
Then, we must have:
2J
= I (0, pi/2) {sin^2(sin x) + cos^2(cos x) + cos^2(sin x) + sin^2(cos x)} dx
= I (0, pi/2) [{sin^2(sin x) + cos^2(sin x) } + {cos^2(cos x) + sin^2(cos x)}]
= I (0, pi/2) (1+1) dx
= I(0, pi/2) (2) dx
= 2* I (0, pi/2) dx
Therefore, we must have:
J
= I(0, pi/2) dx
= pi/2
Consequently, the required definite Integral evaluates to pi/2.
Explanation to Puzzle Answer
