How would you prove $\int _{ 0 }^{ \pi/2 }{f(\sin2x)\sin x dx } $=$\sqrt2\int _{ 0 }^{ \pi/4}{f(\cos2x)\cos x dx }$ ?
Its a problem from IIT-JEE 1990. I've tried it but i'm just going round and round. Give your suggestions please.
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Did
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2By the change of variable $x=\pi/4-u$ and the fact that $\sqrt2\cos x=\cos u+\sin u$, the RHS is $$\int_0^{\pi/4}f(\sin 2u)(\cos u+\sin u)du$$ hence it suffices to show that $$\int_{\pi/4}^{\pi/2}f(\sin 2x)\sin xdx=\int_0^{\pi/4}f(\sin 2u)\cos udu.$$ Now, use the change of variable $x=\pi/2-v$ in the LHS, with $\sin 2x=\sin 2v$ and $\sin x=\cos v$, to conclude. – Did Oct 23 '15 at 19:39
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\begin{align*} \int_0^{\pi/2}f(\sin2x)\sin x\, dx &= \int_{-\pi/4}^{\pi/4} f(\cos 2y)\sin(y+\pi/4) dy \qquad \mbox{ (by } x=y+\pi/4)\\ &=\frac{\sqrt{2}}{2}\int_{-\pi/4}^{\pi/4} f(\cos 2y)(\sin y + \cos y) dy\\ &=\frac{\sqrt{2}}{2}\int_{-\pi/4}^{\pi/4} f(\cos 2y) \cos y\, dy \qquad ( f(\cos 2y) \sin y \mbox { is an odd function})\\ &=\sqrt{2}\int_{0}^{\pi/4} f(\cos 2y) \cos y\, dy \qquad ( f(\cos 2y) \cos y \mbox { is an even function}) \end{align*}
Gordon
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@SanchayanDutta: the integral of an odd function over a symmetric interval is zero. – Gordon Oct 23 '15 at 19:47