The Notes: `int_(pi/4)^(pi/2) cot^5(x) csc^3(x) dx` Evaluate the integral

Sunday, October 18, 2009

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{5}}{\left({x}\right)}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{5}}{\left({x}\right)}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

To solve, apply the Pythagorean identity $\displaystyle{1}+{{\cot}}^{{2}}{\left({x}\right)}={{\csc}}^{{2}}{\left({x}\right)}$ to express the integral in the form $\displaystyle\int{{u}}^{{n}}{d}{u}.$

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

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{2}}{\left({x}\right)}-{1}\right)}{{\cot}}^{{3}}{\left({x}\right)}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{5}}{\left({x}\right)}{{\cot}}^{{3}}{\left({x}\right)}-{{\csc}}^{{3}}{\left({x}\right)}{{\cot}}^{{3}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{5}}{\left({x}\right)}{\cot{{\left({x}\right)}}}{{\cot}}^{{2}}{\left({x}\right)}-{{\csc}}^{{3}}{\left({x}\right)}{\cot{{\left({x}\right)}}}{{\cot}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{5}}{\left({x}\right)}{\cot{{\left({x}\right)}}}{\left({{\csc}}^{{2}}{\left({x}\right)}-{1}\right)}-{{\csc}}^{{3}}{\left({x}\right)}{\cot{{\left({x}\right)}}}{\left({{\csc}}^{{2}}{\left({x}\right)}-{1}\right)}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{7}}{\left({x}\right)}{\cot{{\left({x}\right)}}}-{{\csc}}^{{5}}{\left({x}\right)}{\cot{{\left({x}\right)}}}-{{\csc}}^{{5}}{\left({x}\right)}{\cot{{\left({x}\right)}}}+{{\csc}}^{{3}}{\left({x}\right)}{\cot{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left({{\csc}}^{{7}}{\left({x}\right)}{\cot{{\left({x}\right)}}}-{2}{{\csc}}^{{5}}{\left({x}\right)}{\cot{{\left({x}\right)}}}+{{\csc}}^{{3}}{\left({x}\right)}{\cot{{x}}}\right)}{\left.{d}{x}\right.}$

To take the integral, apply u-substitution method. So, let u be:

u=csc(x) dx

Then, differentiate u.

du = - csc(x) cot(x) dx

To be able to apply this, factor out -csc(x) cot(x).

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left(-{{\csc}}^{{6}}{\left({x}\right)}+{2}{{\csc}}^{{4}}{\left({x}\right)}-{{\csc}}^{{2}}{\left({x}\right)}\right)}{\left(-{\csc{{\left({x}\right)}}}{\cot{{\left({x}\right)}}}{\left.{d}{x}\right.}\right)}$

Then, determine the value of u when x=pi/2 and x=pi/4.

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

$\displaystyle{u}={\csc{{\left(\frac{\pi}{{2}}\right)}}}={1}$

$\displaystyle{u}={\csc{{\left(\frac{\pi}{{4}}\right)}}}=\sqrt{{2}}$

Expressing the integral in terms of u, the integral becomes:

$\displaystyle={\int_{\sqrt{{2}}}^{{1}}}{\left(-{{u}}^{{6}}+{2}{{u}}^{{4}}-{{u}}^{{2}}\right)}{d}{u}$

$\displaystyle={{\left(-\frac{{{u}}^{{7}}}{{7}}+\frac{{{2}{{u}}^{{5}}}}{{5}}-\frac{{{u}}^{{3}}}{{3}}\right)}_{\sqrt{{2}}}^{{1}}}$

$\displaystyle={\left(-\frac{{1}}{{7}}+\frac{{2}}{{5}}-\frac{{1}}{{3}}\right)}-{\left(-\frac{{{\left(\sqrt{{2}}\right)}}^{{7}}}{{7}}+\frac{{{2}{{\left(\sqrt{{2}}\right)}}^{{5}}}}{{5}}-\frac{{{\left(\sqrt{{2}}\right)}}^{{3}}}{{3}}\right)}$

$\displaystyle={0.2201}$

Therefore, $\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{5}}{\left({x}\right)}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}={0.2201}$ .

The Notes

Sunday, October 18, 2009

$\displaystyle{\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\cot}}^{{5}}{\left({x}\right)}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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