Prove that $\int_{0}^{\dfrac{\pi }{2}}{f\left( \sin 2x \right)\sin xdx}=\int_{0}^{\dfrac{\pi }{4}}{f\left( \cos 2x \right)\cos xdx}$ for all f.

Hint: Start from the LHS. Use the fact that $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$. Hence prove that $\int_{0}^{\pi /2}{f\left( \sin 2x \right)\sin xdx}=\int_{0}^{\pi /2}{f\left( \sin 2x \right)\dfrac{\left( \sin x+\cos x \right)}{2}dx}$ . Use the fact that if $f\left( 2a-x \right)=f\left( x \right)$, then $\int_{0}^{2a}{f\left( x \right)dx}=2\int_{0}^{a}{f\left( x \right)dx}$ . Hence prove that $\int_{0}^{\pi /2}{f\left( \sin 2x \right)\sin xdx}=\int_{0}^{\pi /4}{f\left( \sin 2x \right)\left( \sin x+\cos x \right)}$ . Use the fact that $\sin x+\cos x=\sqrt{2}\left( \cos \left( x-\dfrac{\pi }{4} \right) \right)$ and hence prove that $\int_{0}^{\pi /2}{f\left( \sin 2x \right)\sin xdx}=\sqrt{2}\int_{0}^{\pi /4}{f\left( \sin 2x \right)\cos \left( x-\dfrac{\pi }{4} \right)dx}$ . Finally, put $x-\dfrac{\pi }{4}=t$ and hence prove the given result.

Complete step-by-step solution:
Let us take the LHS as $I=\int_{0}^{\pi /2}{f\left( \sin 2x \right)\sin xdx}\ \ \ \ \left( i \right)$
We know that $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$
Hence, we have
$\begin{align}
  & I=\int_{0}^{\pi /2}{f\left( \sin 2\left( \dfrac{\pi }{2}-x \right) \right)\sin \left( \dfrac{\pi }{2}-x \right)dx} \\
 & =\int_{0}^{\pi /2}{f\left( \sin \left( \pi -2x \right) \right)\sin \left( \dfrac{\pi }{2}-x \right)} \\
\end{align}$
We know that $\sin \left( \pi -x \right)=\sin x$ and $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$
Hence, we have
$I=\int_{0}^{\pi /2}{f\left( \sin 2x \right)\cos xdx}\text{ }\left( ii \right)$
Adding equation (i) and equation (ii), we get
$2I=\int_{0}^{\pi /2}{f\left( \sin 2x \right)\left( \sin x+\cos x \right)dx}$
We know that if $f\left( x \right)=f\left( 2a-x \right)$, then
Observe that $f\left( \sin 2x \right)\left( \sin x+\cos x \right)=f\left( \sin 2\left( \dfrac{\pi }{2}-x \right) \right)\left( \sin \left( \dfrac{\pi }{2}-x \right)+\cos \left( \dfrac{\pi }{2}-x \right) \right)$
Hence, we have
$2I=2\int_{0}^{\pi /4}{f\left( \sin 2x \right)\left( \sin x+\cos x \right)dx}$
Hence, we have
$I=\int_{0}^{\pi /4}{f\left( \sin 2x \right)\left( \sin x+\cos x \right)}dx$
Multiplying and dividing by $\sqrt{2}$, we get
$I=\sqrt{2}\int_{0}^{\pi /4}{f\left( \sin 2x \right)\left( \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x \right)}dx$
We know that $\cos \left( \dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$
Hence, we have
$I=\sqrt{2}\int_{0}^{\pi /4}{f\left( \sin 2x \right)\left( \cos x\cos \dfrac{\pi }{4}+\sin x\sin \dfrac{\pi }{4} \right)}dx$
We know that $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$
Hence, we have
$I=\sqrt{2}\int_{0}^{\pi /4}{f\left( \sin 2x \right)\cos \left( \dfrac{\pi }{4}-x \right)}dx$
Put $\dfrac{\pi }{4}-x=t$, we have dx = -dt
When x = 0, $t =\dfrac{\pi }{4}$
When $x=\dfrac{\pi }{4}, t=0$
Hence, we have
$\begin{align}
  & I=\sqrt{2}\left( -\int_{\dfrac{\pi }{4}}^{0}{f\left( \sin \left( 2\left( \dfrac{\pi }{4}-t \right) \right) \right)\cos tdt} \right) \\
 & =\sqrt{2}\int_{0}^{\pi /4}{f\left( \cos 2t \right)\cos tdt} \\
\end{align}$
We know that the change of variable does not affect a definite integral.
Hence, we have
$I=\int_{0}^{\dfrac{\pi }{4}}{f\left( \cos 2x \right)\cos xdx}$
Q.E.D

Note: The identities $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx},\int_{-a}^{a}{f\left( x \right)=\int_{0}^{a}{\left( f\left( x \right)+f\left( -x \right) \right)}}$ and $\int_{0}^{2a}{f\left( x \right)d}x=\int_{0}^{a}{\left( f\left( x \right)+f\left( 2a-x \right) \right)}$ are very important in solving definite integrals. One should consider the use of one of these formulas while solving the question as these make the problem very easy.