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Evaluate \( \int_{0}^{\pi / 2} \sqrt{\sin \theta} d \theta \int_{0}^{\pi / 2} \sqrt{\cos \theta} d \theta \). |
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Answer» please answer this question fast i have my exam tomorrow. \(\int\limits_0^{\pi/2}\sqrt{sin\theta}d\theta\int\limits_0^{\pi/2}\sqrt{cos\theta}d\theta\) \(\int\limits_0^{\pi/2}\sqrt{sin^{1/2}\theta}d\theta\int\limits_0^{\pi/2}\sqrt{cos^{1/2}\theta}d\theta\) \(=\cfrac{\Gamma\left(\frac{\frac12+1}{2}\right)\Gamma(\frac{0+1}2)}{2\Gamma\left(\frac{\frac12+0+2}2\right)}\) \(\times\cfrac{\Gamma\left(\frac{0+1}2\right)\Gamma\left(\frac{\frac12+1}{2}\right)}{2\Gamma\left(\frac{0+\frac12+2}2\right)}\) \((\because\int\limits_0^{\pi/2}sin^m\theta cos^n \theta d\theta= \cfrac{\Gamma(\frac{m+1}2)\Gamma(\frac{n+1}2)}{2\Gamma}(\frac{m+n+2}2))\) \(= \cfrac{\Gamma(\frac34)\Gamma(\frac12)\Gamma(\frac12)\Gamma(\frac34)}{4\Gamma(\frac54)\Gamma(\frac54)}\) \(= \cfrac{\sqrt\pi\times\sqrt{\pi}\,\Gamma(\frac34)\Gamma(\frac34)}{4\Gamma(\frac54)\Gamma(\frac54)}\) \((\because\Gamma(1/2)=\sqrt{\pi})\) \(= \cfrac{\pi}4\frac{[\Gamma(\frac34)]^2}{[\Gamma(\frac54)]^2}\) \(=\frac{\pi}4\left(\cfrac{\Gamma(\frac34)}{\Gamma(\frac54)}\right)^2\) |
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