question_answer

Show that \[\int_{0}^{\pi /2}{f(\sin 2x)\sin \,x\,\,dx}\] \[=\sqrt{2}\int_{0}^{\pi /4}{f(cos2x)cos\,x\,\,dx}\]

Answer:

Let \[l=\int_{0}^{\pi /2}{f(\sin \,\,2x)\sin \,x\,dx}\]                          ?(i) \[\Rightarrow \] \[l=\int_{0}^{\pi /2}{f\left\{ \sin 2\left( \frac{\pi }{2}-x \right) \right\}\sin \left( \frac{\pi }{2}-x \right)}\,dx\] \[\left[ \because \,\,\int_{a}^{b}{f(x)dx=\int_{a}^{b}{(a+b\,-x)\,dx}} \right]\] \[\Rightarrow \] \[l=\int_{0}^{\pi /2}{f\{\sin (\pi -2x)\}\cos \,x\,dx}\] \[\Rightarrow \] \[l=\int_{0}^{\pi /2}{f(\sin 2x)\cos \,x\,dx}\] On adding Eq. (i) and (ii) we get  \[2l=\int_{0}^{\pi /4}{f(\sin \,2x)}\cdot (\sin \,x+\cos \,x)\,dx\] \[\Rightarrow \] \[2l=2\int_{0}^{\pi /4}{f(\sin \,2x)\cdot (\sin \,x+\cos \,x)\,dx}\] \[\left[ \because \,\,\int_{0}^{2a}{f(x)\,dx}=\int_{0}^{a}{2f(x)\,dx},if\,\,f(2a-x)=f(x) \right]\] \[\Rightarrow \] \[2\,\,l=2\sqrt{2}\int_{0}^{\pi /4}{f(\sin \,2x)\left( \frac{1}{\sqrt{2}}\sin \,x \right.}\,\]                                                 \[\left. +\frac{1}{\sqrt{2}}\cos \,x \right)\,dx\] \[\Rightarrow \] \[2\,\,l=2\sqrt{2}\int_{0}^{\pi /4}{f(\sin \,2x)}\left( \cos \frac{\pi }{4}\sin \,x\, \right.\]                                               \[\left. +\sin \frac{\pi }{4}\cos \,x \right)\,dx\] \[\Rightarrow \] \[2l=2\sqrt{2}\int_{0}^{\pi /4}{f(\sin \,2x)}\sin \left( x+\frac{\pi }{4} \right)dx\]            \[[\because \,\,sinA\,cosB+cosA\,sinB=sin(A+B)]\] \[\Rightarrow \] \[2l=2\sqrt{2}\int_{0}^{\pi /4}{f\left\{ \sin 2\left( \frac{\pi }{4}-x \right) \right\}}\]                                                 \[\sin \left( \frac{\pi }{4}-x+\frac{\pi }{4} \right)\,dx\]             \[\left[ \because \,\,\int_{a}^{b}{f(x)\,dx=\int_{a}^{b}{f(a+b-x)\,dx}} \right]\] \[\Rightarrow \] \[2l=2\sqrt{2}\int_{0}^{\pi /4}{f\left\{ \sin \left( \frac{\pi }{2}-2x \right) \right\}}\]                                                 \[\sin \left( \frac{\pi }{2}-x \right)\,dx\] \[\Rightarrow \] \[2l=2\sqrt{2}\int_{0}^{\pi /4}{f(\cos \,2x)\cos \,x\,dx}\]                         \[\left[ \because \,\,\sin \left( \frac{\pi }{2}-\theta  \right)=\cos \theta  \right]\] \[\Rightarrow \] \[l=\sqrt{2}\int_{0}^{\pi /4}{f(\cos \,2x)\,\cos \,x\,dx}\] \[\therefore \,\int_{0}^{\pi /2}{f(\sin \,2x)\sin \,x\,dx=\sqrt{2}\int_{0}^{\pi /4}{f(\cos \,2x)\cos \,x\,dx}}\]                                              Hence proved.