A) \[\pi \int_{0}^{\pi }{f(\sin x)}dx\]
B) \[\frac{\pi }{2}\int_{0}^{\pi /2}{f(\sin x)}dx\]
C) \[\pi \int_{0}^{\pi /2}{f(\cos x)}dx\]
D) \[\pi \int_{0}^{\pi /2}{f(\cos x)}dx\]
Correct Answer: C
Solution :
Let \[l=\int_{0}^{\pi }{xf(\sin x)dx}\] ...(i) \[\Rightarrow \] \[l=\,\int_{0}^{\pi }{(\pi -x)f[\sin (\pi -x)dx}\] \[\Rightarrow \] \[l=\,\int_{0}^{\pi }{(\pi -x)f(\sin x)dx}\] ...(ii) On adding Eqs. (i) and (ii), we get \[2l=\int_{0}^{\pi }{\pi \,f(\sin x)dx}\,\,\Rightarrow \,\,\,l=\frac{\pi }{2}\,\int_{0}^{\pi }{f(\sin x)dx}\] \[\Rightarrow \]\[l=\pi \int_{0}^{\pi }{f(\sin x)dx}\] \[\Rightarrow \]\[l=\pi \int_{0}^{\pi /2}{f\left[ \sin \left( \frac{\pi }{2}-x \right)dx=\pi \int\limits_{0}^{\pi /2}{f(\cos x)dx} \right]}\] [\[\because f(\cos x)\]is an even function]
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