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(i) \( \int_{0}^{\pi} \theta \sin ^{2} \theta \cos \theta d \theta \) |
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Answer» \(\int_0^\pi \theta\,sin^2\theta\,cos\theta\,d\theta\) = \(\int_0^\pi \theta\,(1-cos^2\theta)\,cos\theta\,d\theta\) = \(\int_0^\pi \theta\,cos\theta\,d\theta\) - \(\int_0^\pi \theta\,cos^3\theta\,d\theta\) = \([\theta\int cos\theta\,d\theta]_0^{\pi}\) - \(\int_0^{\pi}(1.\int cos^3\theta\,d\theta)\) = \([\theta\,sin\theta]_0^{\pi}\) - \(\int_0^\pi \theta\,sin\theta\,d\theta\) - \([\theta(sin\theta - \frac{sin^3\theta}{3})]_0^{\pi}\) + \(\int_0^\pi(sin\theta - \frac{sin^3\theta}{3})d\theta\) (∵\(\int cos^3\theta\,d\theta\) = \(\int (1-sin^2\theta)cos\theta\,d\theta\) = \(\int cos\theta\,d\theta\) - \(\int sin^2\theta\,cos\theta\,d\theta\) = \(sin\theta-\frac{sin^3\theta}{3}\)) = (π sinπ - 0) + \([cos\theta]_0^{\pi}\) - (π(sinπ - \(\frac{sin^3\pi}{3}\)) - 0) - \([cos\theta]_0^{\pi}\) - \(\frac{2}{3}\)\(\int_0^{\pi/2} sin^3\theta\,d\theta\) = 0 - 0 - \(\frac{2}{3}\)\(\frac{\Gamma(\frac{3+1}{2})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{2})}\) (\(\int_0^{\pi/2} sin^m\theta\,cos^n \theta d\theta\) = \(\frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+1}{2})}\)) = \(\frac{2}{3}\)\(\frac{\Gamma(2)\Gamma(\frac{1}{2})}{2.\frac{3}{2}.\frac{1}{2}\Gamma(\frac{1}{2})}\) (∵ \(\Gamma(n+1)=n\Gamma(n)\)) = \(-\frac{2}{3}\times \frac{2}{3}\times 1!\) (\(\Gamma(2) = 1!=1\)) = \(\frac{-4}{9}\). |
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