# How do you evaluate int (sin^3x)^(1/2)(cosx)^(1/2) from [0,pi/2]?

Jan 22, 2018

${\int}_{0}^{\frac{\pi}{2}} {\left(\sin x\right)}^{\frac{3}{2}} \cdot {\left(\cos x\right)}^{\frac{1}{2}} \cdot \mathrm{dx} = \frac{\pi \sqrt{2}}{8}$

#### Explanation:

$I = {\int}_{0}^{\frac{\pi}{2}} {\left(\sin x\right)}^{\frac{3}{2}} \cdot {\left(\cos x\right)}^{\frac{1}{2}} \cdot \mathrm{dx}$

=${\int}_{0}^{\frac{\pi}{2}} {\left(\sin \frac{x}{\cos} x\right)}^{\frac{3}{2}} \cdot {\left(\cos x\right)}^{2} \cdot \mathrm{dx}$

=${\int}_{0}^{\frac{\pi}{2}} {\left(\tan x\right)}^{\frac{3}{2}} \cdot {\left(\cos x\right)}^{2} \cdot \mathrm{dx}$

=${\int}_{0}^{\frac{\pi}{2}} {\left(\tan x\right)}^{\frac{3}{2}} \cdot {\left(\sec x\right)}^{2} / {\left(\sec x\right)}^{4} \cdot \mathrm{dx}$

=$2 {\int}_{0}^{\frac{\pi}{2}} {\left(\tan x\right)}^{2} / {\left[{\left(\tan x\right)}^{2} + 1\right]}^{2} \cdot {\left(\sec x\right)}^{2} / \left(2 \sqrt{\tan x}\right) \cdot \mathrm{dx}$

After using $y = \sqrt{\tan x}$ and $\mathrm{dy} = {\left(\sec x\right)}^{2} / \left(2 \sqrt{\tan x}\right) \cdot \mathrm{dx}$ transforms, $I$ became

$I = {\int}_{0}^{\infty} \frac{2 {y}^{4} \cdot \mathrm{dy}}{{y}^{4} + 1} ^ 2$

=$\frac{1}{2} {\int}_{0}^{\infty} y \cdot \frac{4 {y}^{3} \cdot \mathrm{dy}}{{y}^{4} + 1} ^ 2$

=$\frac{1}{2} \cdot {\left[y \cdot - \frac{1}{{y}^{4} + 1}\right]}_{0}^{\infty} - \frac{1}{2} {\int}_{0}^{\infty} - \frac{1}{{y}^{4} + 1} \cdot \mathrm{dy}$

=$- \frac{1}{2} \cdot {\left[\frac{y}{{y}^{4} + 1}\right]}_{0}^{\infty} + \frac{1}{2} {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{4} + 1}$

=$\frac{1}{2} {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{4} + 1}$

After setting $J = {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{4} + 1}$, $I$ must be equal to $\frac{J}{2}$

After using $y = \frac{1}{z}$ and $\mathrm{dy} = - \frac{\mathrm{dz}}{z} ^ 2$ transforms, $J$ became

$J = {\int}_{\infty}^{0} \frac{- \frac{\mathrm{dz}}{z} ^ 2}{{\left(\frac{1}{z}\right)}^{4} + 1}$

=${\int}_{\infty}^{0} \frac{- {z}^{2} \cdot \mathrm{dz}}{{z}^{4} + 1}$

=${\int}_{0}^{\infty} \frac{{z}^{2} \cdot \mathrm{dz}}{{z}^{4} + 1}$

=${\int}_{0}^{\infty} \frac{{y}^{2} \cdot \mathrm{dy}}{{y}^{4} + 1}$

After collecting 2 integrals,

$2 J = {\int}_{0}^{\infty} \frac{\left({y}^{2} + 1\right) \cdot \mathrm{dy}}{{y}^{4} + 1}$

=$\frac{1}{2} {\int}_{0}^{\infty} \frac{\left(2 {y}^{2} + 2\right) \cdot \mathrm{dy}}{{y}^{4} + 1}$

=$\frac{1}{2} {\int}_{0}^{\infty} \frac{\left(2 {y}^{2} + 2\right) \cdot \mathrm{dy}}{\left({y}^{2} + \sqrt{2} \cdot y + 1\right) \cdot \left({y}^{2} - \sqrt{2} \cdot y + 1\right)}$

=$\frac{1}{2} {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{2} + \sqrt{2} \cdot y + 1} + \frac{1}{2} {\int}_{0}^{\infty} \frac{\mathrm{dy}}{{y}^{2} - \sqrt{2} \cdot y + 1}$

=$\frac{1}{2} {\int}_{0}^{\infty} \frac{2 \mathrm{dy}}{2 {y}^{2} + 2 \sqrt{2} \cdot y + 2} + \frac{1}{2} {\int}_{0}^{\infty} \frac{2 \mathrm{dy}}{2 {y}^{2} - 2 \sqrt{2} \cdot y + 2}$

=$\frac{\sqrt{2}}{2} {\int}_{0}^{\infty} \frac{\sqrt{2} \mathrm{dy}}{{\left(\sqrt{2} \cdot y + 1\right)}^{2} + 1} + \frac{\sqrt{2}}{2} {\int}_{0}^{\infty} \frac{\sqrt{2} \mathrm{dy}}{{\left(\sqrt{2} \cdot y - 1\right)}^{2} + 1}$

=$\frac{\sqrt{2}}{2} \cdot {\left[\arctan \left(\sqrt{2} \cdot y + 1\right)\right]}_{0}^{\infty} + \frac{\sqrt{2}}{2} \cdot {\left[\arctan \left(\sqrt{2} \cdot y - 1\right)\right]}_{0}^{\infty}$

=$\frac{\pi \sqrt{2}}{2}$

Thus,

$I = \frac{J}{2} = \frac{\pi \sqrt{2}}{8}$