The Notes: `int csc^4(x) cot^6(x) dx` Evaluate the integral

$\displaystyle\int{{\csc}}^{{4}}{\left({x}\right)}{{\cot}}^{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$

Rewrite the integrand by applying the identity, $\displaystyle{1}+{{\cot}}^{{2}}{\left({x}\right)}={{\csc}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int{{\left({1}+{{\cot}}^{{2}}{\left({x}\right)}\right)}}^{{2}}{{\cot}}^{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$

Apply integral substitution,

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

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

$\displaystyle\Rightarrow{\left.{d}{x}\right.}={\left(-\frac{{1}}{{{{\csc}}^{{2}}{\left({x}\right)}}}\right)}{d}{u}=-\frac{{1}}{{{1}+{{u}}^{{2}}}}{d}{u}$

$\displaystyle=\int{{\left({1}+{{u}}^{{2}}\right)}}^{{2}}{{u}}^{{6}}{\left(-\frac{{1}}{{{1}+{{u}}^{{2}}}}\right)}{d}{u}$

$\displaystyle=-\int{\left({1}+{{u}}^{{2}}\right)}{{u}}^{{6}}{d}{u}$

$\displaystyle=-\int{\left({{u}}^{{6}}+{{u}}^{{8}}\right)}{d}{u}$

$\displaystyle=-{\left(\int{{u}}^{{6}}{d}{u}+\int{{u}}^{{8}}{d}{u}\right)}$

$\displaystyle=-{\left(\frac{{{u}}^{{7}}}{{7}}+\frac{{{u}}^{{9}}}{{9}}\right)}$

$\displaystyle=-\frac{{1}}{{7}}{{u}}^{{7}}-\frac{{1}}{{9}}{{u}}^{{9}}$

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

$\displaystyle=-\frac{{1}}{{7}}{{\cot}}^{{7}}{\left({x}\right)}-\frac{{1}}{{9}}{{\cot}}^{{9}}{\left({x}\right)}$

Add a constant C to the solution,

$\displaystyle=-\frac{{1}}{{7}}{{\cot}}^{{7}}{\left({x}\right)}-\frac{{1}}{{9}}{{\cot}}^{{9}}{\left({x}\right)}+{C}$

The Notes

Thursday, December 30, 2010

$\displaystyle\int{{\csc}}^{{4}}{\left({x}\right)}{{\cot}}^{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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