Notes: `int 1/(2-3sin(theta)) d theta` Find or evaluate the integral

$\displaystyle\int\frac{{1}}{{{2}-{3}{\sin{{\left(\theta\right)}}}}}{d}\theta$

Apply integral substitution: $\displaystyle{u}={\tan{{\left(\frac{\theta}{{2}}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=\frac{{1}}{{2}}{{\sec}}^{{2}}{\left(\frac{\theta}{{2}}\right)}{d}\theta$

Use the trigonometric identity: $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle{{\sec}}^{{2}}{\left(\frac{\theta}{{2}}\right)}={1}+{{\tan}}^{{2}}{\left(\frac{\theta}{{2}}\right)}$

$\displaystyle{{\sec}}^{{2}}{\left(\frac{\theta}{{2}}\right)}={1}+{{u}}^{{2}}$

$\displaystyle{d}{u}=\frac{{1}}{{2}}{\left({1}+{{u}}^{{2}}\right)}{d}\theta$

$\displaystyle{d}\theta=\frac{{2}}{{{1}+{{u}}^{{2}}}}{d}{u}$

From integral substitution: $\displaystyle{u}={\tan{{\left(\frac{\theta}{{2}}\right)}}}$

$\displaystyle\Rightarrow{\sin{{\left(\frac{\theta}{{2}}\right)}}}=\frac{{u}}{\sqrt{{{{u}}^{{2}}+{1}}}}$

$\displaystyle{\cos{{\left(\frac{\theta}{{2}}\right)}}}=\frac{{1}}{\sqrt{{{{u}}^{{2}}+{1}}}}$

$\displaystyle{\sin{{\left(\theta\right)}}}={2}{\sin{{\left(\frac{\theta}{{2}}\right)}}}{\cos{{\left(\frac{\theta}{{2}}\right)}}}$

$\displaystyle{\sin{{\left(\theta\right)}}}={2}{\left(\frac{{u}}{\sqrt{{{{u}}^{{2}}+{1}}}}\right)}{\left(\frac{{1}}{\sqrt{{{{u}}^{{2}}+{1}}}}\right)}$

$\displaystyle{\sin{{\left(\theta\right)}}}=\frac{{{2}{u}}}{{{{u}}^{{2}}+{1}}}$

Now the integrand can be written as :

$\displaystyle\int\frac{{1}}{{{2}-{3}{\sin{{\left(\theta\right)}}}}}{d}\theta=\int\frac{{1}}{{{2}-{3}{\left(\frac{{{2}{u}}}{{{{u}}^{{2}}+{1}}}\right)}}}{\left(\frac{{2}}{{{1}+{{u}}^{{2}}}}\right)}{d}{u}$

$\displaystyle=\int\frac{{1}}{{\frac{{{2}{\left({{u}}^{{2}}+{1}\right)}-{3}{\left({2}{u}\right)}}}{{{{u}}^{{2}}+{1}}}}}{\left(\frac{{2}}{{{1}+{{u}}^{{2}}}}\right)}{d}{u}$

$\displaystyle=\int\frac{{2}}{{{2}{{u}}^{{2}}+{2}-{6}{u}}}{d}{u}$

$\displaystyle=\int\frac{{2}}{{{2}{\left({{u}}^{{2}}-{3}{u}+{1}\right)}}}{d}{u}$

$\displaystyle=\int\frac{{1}}{{{{u}}^{{2}}-{3}{u}+{1}}}{d}{u}$

Complete the square of the denominator,

$\displaystyle=\int\frac{{1}}{{{{\left({u}-\frac{{3}}{{2}}\right)}}^{{2}}-\frac{{5}}{{4}}}}{d}{u}$

Again apply integral substitution: $\displaystyle{v}={u}-\frac{{3}}{{2}}$

$\displaystyle\Rightarrow{d}{v}={1}{d}{u}$

$\displaystyle=\int\frac{{1}}{{{{v}}^{{2}}-{{\left(\frac{\sqrt{{{5}}}}{{2}}\right)}}^{{2}}}}{d}{v}$

$\displaystyle=\int\frac{{1}}{{-{1}{\left({{\left(\frac{\sqrt{{{5}}}}{{2}}\right)}}^{{2}}-{{v}}^{{2}}\right)}}}{d}{v}$

Take the constant out,

$\displaystyle=-{1}\int\frac{{1}}{{{{\left(\frac{\sqrt{{{5}}}}{{2}}\right)}}^{{2}}-{{v}}^{{2}}}}{d}{v}$

Now use the standard table integral: $\displaystyle\int\frac{{1}}{{{{a}}^{{2}}-{{x}}^{{2}}}}{\left.{d}{x}\right.}=\frac{{1}}{{{2}{a}}}{\ln}{\left|\frac{{{a}+{x}}}{{{a}-{x}}}\right|}+{C}$

$\displaystyle=-{1}{\left(\frac{{1}}{{{2}{\left(\frac{\sqrt{{{5}}}}{{2}}\right)}}}{\ln}{\left|\frac{{\frac{\sqrt{{{5}}}}{{2}}+{v}}}{{\frac{\sqrt{{{5}}}}{{2}}-{v}}}\right|}\right)}+{C}$

$\displaystyle={\left(-\frac{{1}}{\sqrt{{{5}}}}\right)}{\ln}{\left|\frac{{\sqrt{{{5}}}+{2}{v}}}{{\sqrt{{{5}}}-{2}{v}}}\right|}+{C}$

Substitute back $\displaystyle{v}={u}-\frac{{3}}{{2}}$

$\displaystyle={\left(-\frac{{1}}{\sqrt{{{5}}}}\right)}{\ln}{\left|\frac{{\sqrt{{{5}}}+{2}{\left({u}-\frac{{3}}{{2}}\right)}}}{{\sqrt{{{5}}}-{2}{\left({u}-\frac{{3}}{{2}}\right)}}}\right|}+{C}$

$\displaystyle={\left(-\frac{{1}}{\sqrt{{{5}}}}\right)}{\ln}{\left|\frac{{\sqrt{{{5}}}+{2}{u}-{6}}}{{\sqrt{{{5}}}-{2}{u}+{6}}}\right|}+{C}$

Substitute back $\displaystyle{u}={\tan{{\left(\frac{\theta}{{2}}\right)}}}$

$\displaystyle={\left(-\frac{{1}}{\sqrt{{{5}}}}\right)}{\ln}{\left|\frac{{\sqrt{{{5}}}+{2}{\tan{{\left(\frac{\theta}{{2}}\right)}}}-{6}}}{{\sqrt{{{5}}}-{2}{\tan{{\left(\frac{\theta}{{2}}\right)}}}+{6}}}\right|}+{C}$