`int_(pi/4)^(pi/2) cot^3(x) dx` Evaluate the integral

Sunday, July 19, 2015

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{3}}{\left({x}\right)}$

Let's evaluate the indefinite integral by rewriting the integrand as,

$\displaystyle\int{{\cot}}^{{3}}{\left({x}\right)}=\int{\cot{{\left({x}\right)}}}{{\cot}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

Now use the identity: $\displaystyle{{\cot}}^{{2}}{\left({x}\right)}={{\csc}}^{{2}}{\left({x}\right)}-{1}$

$\displaystyle=\int{\cot{{\left({x}\right)}}}{\left({{\csc}}^{{2}}{\left({x}\right)}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left({\cot{{\left({x}\right)}}}{{\csc}}^{{2}}{\left({x}\right)}-{\cot{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int{\cot{{\left({x}\right)}}}{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}-\int{\cot{{\left({x}\right)}}}{\left.{d}{x}\right.}$

Now let's evaluate $\displaystyle\int{\cot{{\left({x}\right)}}}{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ by integral substitution,

Let $\displaystyle{u}={\cot{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=-{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle\int{\cot{{\left({x}\right)}}}{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}=\int{u}{\left(-{d}{u}\right)}$

$\displaystyle=-\int{u}{d}{u}$

$\displaystyle=-\frac{{{u}}^{{2}}}{{2}}$

substitute back $\displaystyle{u}={\cot{{\left({x}\right)}}}$

$\displaystyle=-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left({x}\right)}$

Use the common integral $\displaystyle\int{\cot{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln}{\left|{\sin{{\left({x}\right)}}}\right|}$

$\displaystyle\therefore\int{{\cot}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}=-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left({x}\right)}-{\ln}{\left|{\sin{{\left({x}\right)}}}\right|}+{C}$ , C is a constant

Now let' evaluate the definite integral,

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}={{\left[-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left({x}\right)}-{\ln}{\left|{\sin{{\left({x}\right)}}}\right|}\right\rbrace}_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle={\left[-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{\pi}{{2}}\right)}-{\ln}{\left|{\sin{{\left(\frac{\pi}{{2}}\right)}}}\right|}\right]}-{\left[-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{\pi}{{4}}\right)}-{\ln}{\left|{\sin{{\left(\frac{\pi}{{4}}\right)}}}\right|}\right]}$

$\displaystyle={\left[-\frac{{1}}{{2}}\cdot{0}-{\ln{{\left({1}\right)}}}\right]}-{\left[-\frac{{1}}{{2}}{{\left({1}\right)}}^{{2}}-{\ln{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}}\right]}$

$\displaystyle={\left[{0}\right]}+\frac{{1}}{{2}}+{\ln{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}}$

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

$\displaystyle=\frac{{1}}{{2}}-\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left({1}-{\ln{{\left({2}\right)}}}\right)}$