Question #3cdcc

Nov 13, 2017

$\int \frac{\sin x \cdot \mathrm{dx}}{\sqrt{2 \cdot \left(1 + \cos x\right)}} = - 2 \cos \left(\frac{x}{2}\right) + C = - \sqrt{2 \cdot \left(1 + \cos x\right)} + C$

Explanation:

$\int \frac{\sin x \cdot \mathrm{dx}}{\sqrt{2 \cdot \left(1 + \cos x\right)}}$

=$\int \frac{2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right) \cdot \mathrm{dx}}{\sqrt{4 {\left(\cos \left(\frac{x}{2}\right)\right)}^{2}}}$

=$\int \frac{2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right) \cdot \mathrm{dx}}{2 \cos \left(\frac{x}{2}\right)}$

=$\int \sin \left(\frac{x}{2}\right) \cdot \mathrm{dx}$

=$- 2 \cos \left(\frac{x}{2}\right) + C$

=$- 2 \cdot \sqrt{\frac{1 + \cos x}{2}} + C$

=$- \sqrt{2 \cdot \left(1 + \cos x\right)} + C$

Nov 22, 2017

$\int {\left(\sin x\right)}^{2} / \left(\sin x + 2 \cos x\right) \cdot \mathrm{dx}$=$- \frac{1}{5} \cdot \cos x - \frac{2}{5} \cdot \sin x - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 \tan \left(\frac{x}{2}\right) - 1 - \sqrt{5}}{2 \tan \left(\frac{x}{2}\right) - 1 + \sqrt{5}}\right] + C$

Explanation:

$\int {\left(\sin x\right)}^{2} / \left(\sin x + 2 \cos x\right) \cdot \mathrm{dx}$

After using $u = \tan \left(\frac{x}{2}\right)$, $\cos u = \frac{1 - {u}^{2}}{{u}^{2} + 1}$, $\sin u = \frac{2 u}{{u}^{2} + 1}$ and $\mathrm{dx} = \frac{2 \mathrm{du}}{{u}^{2} + 1}$ transforms, this integral became

$\int {\left[\frac{2 u}{{u}^{2} + 1}\right]}^{2} / \left[2 \frac{u}{{u}^{2} + 1} + 2 \cdot \frac{1 - {u}^{2}}{{u}^{2} + 1}\right] \cdot \frac{2 \mathrm{du}}{{u}^{2} + 1}$

=$\int \frac{- 4 {u}^{2} \cdot \mathrm{du}}{{\left({u}^{2} + 1\right)}^{2} \cdot \left({u}^{2} - u - 1\right)}$

After decomposing integrand into basic fractions,

$\int \frac{- 4 {u}^{2} \cdot \mathrm{du}}{{\left({u}^{2} + 1\right)}^{2} \cdot \left({u}^{2} - u - 1\right)}$

=$\frac{4}{5} \cdot \int \frac{\mathrm{du}}{{u}^{2} + 1} + \frac{4}{5} \cdot \int \frac{\left(u - 2\right) \cdot \mathrm{du}}{{u}^{2} + 1} ^ 2 - \frac{4}{5} \cdot \int \frac{\mathrm{du}}{{u}^{2} - u - 1}$

=$\frac{4}{5} \cdot \arctan u + \frac{2}{5} \cdot \int \frac{2 u \cdot \mathrm{du}}{{u}^{2} + 1} ^ 2 - \frac{8}{5} \cdot \int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2$-$\frac{16}{5} \cdot \int \frac{\mathrm{du}}{4 {u}^{2} - 4 u - 4}$

=$\frac{4}{5} \cdot \arctan u - \frac{2}{5} \cdot {\left({u}^{2} + 1\right)}^{- 1} + C 1 - \frac{8}{5} \cdot \int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2$-$\frac{8}{5} \cdot \int \frac{2 \mathrm{du}}{{\left(2 u - 1\right)}^{2} - 5}$

$I = \int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2$

After taking $u = \tan y$ and $\mathrm{du} = {\left(\sec y\right)}^{2} \cdot \mathrm{dz}$ transormation, $I$ became

$I = \int \frac{{\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{\sec y} ^ 4$

=$\int {\left(\cos y\right)}^{2} \cdot \mathrm{dy}$

=$\int \frac{1}{2} \cdot \left(1 + \cos 2 y\right) \cdot \mathrm{dy}$

=$\frac{1}{2} \cdot y + \frac{1}{4} \cdot \sin 2 y$

After using $u = \tan y$, $y = \arctan u$ and $\sin 2 y = 2 \frac{u}{{u}^{2} + 1}$ inverse transforms, I found

$I = \int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2 = \frac{1}{2} \cdot \arctan u + \frac{1}{2} \cdot \frac{u}{{u}^{2} + 1}$

$J = \int \frac{2 \mathrm{du}}{{\left(2 u - 1\right)}^{2} - 5}$

After taking $2 u - 1 = z$ and $2 \mathrm{du} = \mathrm{dz}$ transforms, $J$ became

$J = \int \frac{\mathrm{dz}}{{z}^{2} - 5}$

=$\int \frac{\mathrm{dz}}{\left(z + \sqrt{5}\right) \cdot \left(z - \sqrt{5}\right)}$

=$\frac{\sqrt{5}}{10} \cdot \int \frac{\mathrm{dz}}{z - \sqrt{5}} - \frac{\sqrt{5}}{10} \cdot \int \frac{\mathrm{dz}}{z + \sqrt{5}}$

=$\frac{\sqrt{5}}{10} \cdot L n \left(z - \sqrt{5}\right) - \frac{\sqrt{5}}{10} \cdot L n \left(z + \sqrt{5}\right)$

=$\frac{\sqrt{5}}{10} \cdot L n \left[\frac{z - \sqrt{5}}{z + \sqrt{5}}\right]$

=$= \frac{\sqrt{5}}{10} \cdot L n \left[\frac{2 u - 1 - \sqrt{5}}{2 u - 1 + \sqrt{5}}\right]$

Hence,

$\int \frac{- 4 {u}^{2} \cdot \mathrm{du}}{{\left({u}^{2} + 1\right)}^{2} \cdot \left({u}^{2} - u - 1\right)}$

=$\frac{4}{5} \cdot \arctan u - \frac{2}{5} \cdot {\left({u}^{2} + 1\right)}^{- 1} + C 1$-$\frac{8}{5} \cdot \left[\frac{1}{2} \cdot \arctan u + \frac{1}{2} \cdot \frac{u}{{u}^{2} + 1}\right]$
-$\frac{8}{5} \cdot \frac{\sqrt{5}}{10} \cdot L n \left[\frac{2 u - 1 - \sqrt{5}}{2 u - 1 + \sqrt{5}}\right]$

=$\frac{4}{5} \cdot a r \tan u - \frac{2}{5} \cdot {\left({u}^{2} + 1\right)}^{- 1} - \frac{4}{5} \arctan u - \frac{4}{5} \cdot \frac{u}{{u}^{2} + 1} - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 u - 1 - \sqrt{5}}{2 u - 1 + \sqrt{5}}\right] + C 1$

=$- \frac{2}{5} \cdot {\left({u}^{2} + 1\right)}^{- 1} - \frac{4}{5} \cdot \frac{u}{{u}^{2} + 1} - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 u - 1 - \sqrt{5}}{2 u - 1 + \sqrt{5}}\right] + C 1$

Thus,

$\int {\left(\sin x\right)}^{2} / \left(\sin x + 2 \cos x\right) \cdot \mathrm{dx}$

=$- \frac{2}{5} \cdot \left[{\left(\tan {\left(\frac{x}{2}\right)}^{2} + 1\right]}^{- 1} - \frac{4}{5} \cdot \tan \frac{\frac{x}{2}}{{\left(\tan \left(\frac{x}{2}\right)\right)}^{2} + 1}\right] - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 \tan \left(\frac{x}{2}\right) - 1 - \sqrt{5}}{2 \tan \left(\frac{x}{2}\right) - 1 + \sqrt{5}}\right] + C 1$

=$- \frac{2}{5} \cdot {\left(\cos \left(\frac{x}{2}\right)\right)}^{2} - \frac{2}{5} \cdot \sin x - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 \tan \left(\frac{x}{2}\right) - 1 - \sqrt{5}}{2 \tan \left(\frac{x}{2}\right) - 1 + \sqrt{5}}\right] + C 1$

=$- \frac{1}{5} \cdot \cos x - \frac{2}{5} \cdot \sin x - \frac{4 \sqrt{5}}{25} \cdot L n \left[\frac{2 \tan \left(\frac{x}{2}\right) - 1 - \sqrt{5}}{2 \tan \left(\frac{x}{2}\right) - 1 + \sqrt{5}}\right] + C$

Note: $C = C 1 - \frac{1}{5}$