IIT-JEE 2003Question 1
If $f$ is an even function then prove that $$\int_0^{\pi/2}f(\cos2x)\cos x\,dx=\sqrt2\int_0^{\pi/4}f(\sin2x)\cos x\,dx.$$
Definite Integration
Answerproved
Solution
Concept. Use $x\to\frac\pi2-x$ ($f$ even) to get a second form with $\sin x$; add, halve the interval by symmetry about $\frac\pi4$, and finally shift by $\frac\pi4$ to turn $\cos2x$ into $\sin2u$. Formulas used. (1) $\cos\big(2(\tfrac\pi2-x)\big)=\cos(\pi-2x)=-\cos2x$, and $f(-t)=f(t)$(2) $\int_0^{2a}g(x)\,dx=2\int_0^ag(x)\,dx$ if $g(2a-x)=g(x)$(3) $\sin x+\cos x=\sqrt2\cos\big(x-\frac\pi4\big)$Step 1 - Second form. Let $L=\int_0^{\pi/2}f(\cos2x)\cos x\,dx$. With $x\to\frac\pi2-x$ and (1), $$L=\int_0^{\pi/2}f(\cos2x)\sin x\,dx.$$ Step 2 - Add. $$2L=\int_0^{\pi/2}f(\cos2x)(\sin x+\cos x)\,dx.$$ Step 3 - Halve. $g(x)=f(\cos2x)(\sin x+\cos x)$ satisfies $g\big(\frac\pi2-x\big)=g(x)$ by (1). By (2) with $a=\frac\pi4$, $$L=\int_0^{\pi/4}f(\cos2x)(\sin x+\cos x)\,dx.$$ Step 4 - Put $x=\frac\pi4-u$. $\cos2x=\cos\big(\frac\pi2-2u\big)=\sin2u$, and by (3) $\sin x+\cos x=\sqrt2\cos u$. Limits $x:0\to\frac\pi4$ give $u:\frac\pi4\to0$, $dx=-du$: $$L=\sqrt2\int_0^{\pi/4}f(\sin2u)\cos u\,du.$$ This is the required identity. Hence proved. Answer: proved