The value of the integral int_0^{pi / 2} frac | vedclass

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(D) Let $I = \int_0^{\pi / 2} \frac{3 \sqrt{\cos 	heta}}{(\sqrt{\cos 	heta}+\sqrt{\sin 	heta})^5} d 	heta$. Using the property $\int_0^a f(x) dx =

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$0.50$

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(D) Let $I = \int_0^{\pi / 2} \frac{3 \sqrt{\cos 	heta}}{(\sqrt{\cos 	heta}+\sqrt{\sin 	heta})^5} d 	heta$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi / 2} \frac{3 \sqrt{\sin 	heta}}{(\sqrt{\sin 	heta}+\sqrt{\cos 	heta})^5} d 	heta$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{3(\sqrt{\cos 	heta} + \sqrt{\sin 	heta})}{(\sqrt{\cos 	heta}+\sqrt{\sin 	heta})^5} d 	heta = \int_0^{\pi / 2} \frac{3}{(\sqrt{\cos 	heta}+\sqrt{\sin 	heta})^4} d 	heta$. Divide numerator and denominator by $\cos^2 	heta$: $2I = 3 \int_0^{\pi / 2} \frac{\sec^2 	heta}{(1+\sqrt{	an 	heta})^4} d 	heta$. Let $1+\sqrt{	an 	heta} = t$,then $\frac{1}{2\sqrt{	an 	heta}} \cdot \sec^2 	heta d 	heta = dt$,so $\sec^2 	heta d 	heta = 2(t-1) dt$. Substituting these: $2I = 3 \int_1^{\infty} \frac{2(t-1)}{t^4} dt = 6 \int_1^{\infty} (t^{-3} - t^{-4}) dt$. $2I = 6 \left[ \frac{t^{-2}}{-2} - \frac{t^{-3}}{-3} ight]_1^{\infty} = 6 \left[ 0 - (-\frac{1}{2} + \frac{1}{3}) ight] = 6 \left( \frac{1}{6} ight) = 1$. Thus,$I = 0.50$.

The value of the integral $\int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$ equals

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