Notes: `int_0^(pi/2) 1/(1+sin(theta)+cos(theta)) d theta` Find or evaluate the integral

Saturday, 20 June 2015

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{1}}{{{1}+{\sin{{\left(\theta\right)}}}+{\cos{{\left(\theta\right)}}}}}{d}\theta$ Find or evaluate the integral

To evaluate the integral problem: $\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{1}}{{{1}+{\sin{{\left(\theta\right)}}}+{\cos{{\left(\theta\right)}}}}}{d}\theta$ , we may apply Weierstrass substitution or tangent half-angle substitution .

This helps to determine the indefinite integral of a rational function in terms of sine and cosine. We let:

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

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

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

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

Plug-in the values to express the integral problem in terms variable "u'.

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

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

$\displaystyle=\int\frac{{1}}{{{\left(\frac{{{1}+{{u}}^{{2}}+{2}{u}+{1}-{{u}}^{{2}}}}{{{1}+{{u}}^{{2}}}}\right)}}}\cdot{\left({2}\ldots\right.}$

This helps to determine the indefinite integral of a rational function in terms of sine and cosine. We let:

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

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

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

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

Plug-in the values to express the integral problem in terms variable "u'.

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

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

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

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

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

$\displaystyle=\int\frac{{{d}{u}}}{{{1}+{u}}}$

From the table of indefinite integration table, we follow the integral formula for rational function as:

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{a}{x}+{b}}}=\frac{{1}}{{a}}{\ln{{\left({a}{x}+{b}\right)}}}$

By comparing " $\displaystyle{a}{x}+{b}$ " with " $\displaystyle{1}+{u}$ or $\displaystyle{1}{u}+{1}$ ", the corresponding values are: $\displaystyle{a}={1}$ and $\displaystyle{b}={1}$ . Then, the integral becomes:

$\displaystyle\int\frac{{{d}{u}}}{{{1}+{u}}}=\frac{{1}}{{1}}{\ln{{\left({1}{u}+{1}\right)}}}$

$\displaystyle={\ln{{\left({u}+{1}\right)}}}$

Plug-in $\displaystyle{u}={\tan{{\left(\frac{{x}}{{2}}\right)}}}$ on $\displaystyle{\ln{{\left({u}+{1}\right)}}}$ , we get:

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{1}}{{{1}+{\sin{{\left(\theta\right)}}}+{\cos{{\left(\theta\right)}}}}}{d}\theta={\ln{{\left({\tan{{\left(\frac{{x}}{{2}}\right)}}}+{1}\right)}}}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

Apply the definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{\ln{{\left({\tan{{\left(\frac{{x}}{{2}}\right)}}}+{1}\right)}}}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}={\ln{{\left({\tan{{\left(\frac{{{\left(\frac{\pi}{{2}}\right)}}}{{2}}\right)}}}+{1}\right)}}}-{\ln{{\left({\tan{{\left(\frac{{0}}{{2}}\right)}}}+{1}\right)}}}$