The Notes: `int_0^(pi/3) tan^5(x) sec^4(x) dx` Evaluate the integral

$\displaystyle{{\tan}}^{{5}}{\left({x}\right)}{{\sec}}^{{4}}{\left({x}\right)}=\frac{{{{\sin}}^{{5}}{\left({x}\right)}}}{{{{\cos}}^{{9}}{\left({x}\right)}}}=\frac{{{\left({1}-{{\cos}}^{{2}}{\left({x}\right)}\right)}}^{{2}}}{{{{\cos}}^{{9}}{\left({x}\right)}}}\cdot{\sin{{\left({x}\right)}}}.$

Make a substitution $\displaystyle{y}={\cos{{\left({x}\right)}}},$ then $\displaystyle{\left.{d}{y}\right.}=-{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}$ and the indefinite integral is equal to

$\displaystyle-\int\frac{{{\left({1}-{{y}}^{{2}}\right)}}^{{2}}}{{{y}}^{{9}}}{\left.{d}{y}\right.}=-\int{\left({{y}}^{{-{9}}}-{2}{{y}}^{{-{7}}}+{{y}}^{{-{5}}}\right)}{\left.{d}{y}\right.}=$

$\displaystyle=\frac{{1}}{{8}}{{y}}^{{-{8}}}-\frac{{1}}{{3}}{{y}}^{{-{6}}}+\frac{{1}}{{4}}{{y}}^{{-{4}}}+{C}.$

And the definite integral is equal to

$\displaystyle{\left(\frac{{1}}{{8}}{{y}}^{{-{8}}}-\frac{{1}}{{3}}{{y}}^{{-{6}}}+\frac{{1}}{{4}}{{y}}^{{-{4}}}\right)}{{\mid}_{{{y}={\cos{{\left({0}\right)}}}={1}}}^{{{y}={\cos{{\left(\frac{\pi}{{3}}\right)}}}=\frac{{1}}{{2}}}}}=$

$\displaystyle=\frac{{1}}{{8}}{\left({{2}}^{{8}}-{1}\right)}-\frac{{1}}{{3}}{\left({{2}}^{{6}}-{1}\right)}+\frac{{1}}{{4}}{\left({{2}}^{{4}}-{1}\right)}=\frac{{255}}{{8}}-{21}+\frac{{30}}{{8}}=\frac{{117}}{{8}}.$

The Notes

Tuesday, April 8, 2014

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{3}}}}}{{\tan}}^{{5}}{\left({x}\right)}{{\sec}}^{{4}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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