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(ii) \( \int_{0}^{\pi} x \sin ^{6} x \cos ^{4} x d x \) |
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Answer» Let I = \(\int\limits_0^\pi x\,sin^6x\,cos^4x\,dx\) ....(1) ∴ I = \(\int\limits_0^\pi (\pi -x)\,sin^6(\pi-x)\,cos^4(\pi-x)\,dx\) ....(2) ∴ 2I = \(\int\limits_0^\pi \pi\,sin^6x\,cos^4x\,dx\) (By adding (1) & (2)) ⇒ I = \(\frac{\pi}{2}\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\) = \(\frac{\pi}{2}\) x 2\(\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\) = π \(\frac{\Gamma(\frac{6+1}{2}).\Gamma(\frac{4+1}{2})}{2\Gamma(\frac{6+4+2}{2})}\)(By gamma function) = \(\frac{\pi}{2}\) \(\frac{\Gamma(\frac{7}{2}).\Gamma(\frac{5}{2})}{\Gamma(6)}\) = \(\frac{\pi}{2}\)\(\frac{\frac{5}{2}.\frac{3}{2}.\Gamma(\frac{1}{2})\times \frac{3}{2}.\frac{1}{2}.\Gamma(\frac{1}{2})}{5!}\) (\(\Gamma(n) = (n-1)!\) n ∈ N & \(\Gamma(n+1)=n\Gamma(n)\)) = \(\frac{\pi}{2}\) x \(\frac{45}{32}\) x \(\frac{1}{120}\) √π x √π (∵ \(\Gamma(\frac{1}{2})= \sqrt {\pi}\)) = \(\frac{3\pi^2}{512}\) |
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