The value of the integral $\int_{0}^{\pi/2} \frac{3\sqrt{\cos\theta}}{\left(\sqrt{\cos\theta}+\sqrt{\sin\theta}\right)^{5}}\,d\theta$ equals ___

The value of the integral ∫₀^(π/2) 3√cosθ/((√cosθ+√sinθ)⁵) dθ equals ___

$I = \displaystyle\int_{0}^{\pi/2} \dfrac{3\sqrt{\cos\theta}}{\left(\sqrt{\cos\theta}+\sqrt{\sin\theta}\right)^{5}}\,d\theta$

$\text{Find } I$

$1.\ \text{substitute } t = \tan\theta$

$2.\ \text{simplify the integrand}$

$3.\ \text{evaluate with a further substitution}$

$t = \tan\theta, \quad d\theta = \dfrac{dt}{1+t^2}$

$\sqrt{\cos\theta} = (1+t^2)^{-1/4}, \quad \sqrt{\sin\theta} = \sqrt{t}\,(1+t^2)^{-1/4}$

$\sqrt{\cos\theta}+\sqrt{\sin\theta} = (1+\sqrt{t})(1+t^2)^{-1/4}$

$I = \displaystyle\int_{0}^{\infty} \dfrac{3}{(1+\sqrt{t})^{5}}\,dt$

$u = \sqrt{t}, \quad dt = 2u\,du$

$I = 6\displaystyle\int_{0}^{\infty} \dfrac{u}{(1+u)^{5}}\,du$

$\int_{0}^{\infty} \dfrac{u}{(1+u)^{5}}du = \int_{1}^{\infty}\dfrac{v-1}{v^{5}}dv = \int_{1}^{\infty}\left(v^{-4}-v^{-5}\right)dv$

$= \left[-\dfrac{1}{3v^{3}}+\dfrac{1}{4v^{4}}\right]_{1}^{\infty} = \dfrac{1}{3}-\dfrac{1}{4} = \dfrac{1}{12}$

$I = 6\cdot\dfrac{1}{12} = \dfrac{1}{2}$