Notes: `int tan^5(2x)sec^4(2x) dx` Find the indefinite integral

Wednesday, 12 March 2014

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$

Let's apply integral substitution: $\displaystyle{u}={2}{x}$

$\displaystyle{\left({d}{u}\right)}={2}{\left.{d}{x}\right.}$

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}=\int{{\tan}}^{{5}}{\left({u}\right)}{{\sec}}^{{4}}{\left({u}\right)}\frac{{{d}{u}}}{{2}}$

Take the constant out and rewrite the integral as,

$\displaystyle=\frac{{1}}{{2}}\int{{\sec}}^{{2}}{\left({u}\right)}{{\sec}}^{{2}}{\left({u}\right)}{{\tan}}^{{5}}{\left({u}\right)}{d}{u}$

Now use the trigonometric identity : $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({1}+{{\tan}}^{{2}}{\left({u}\right)}\right)}{{\sec}}^{{2}}{\left({u}\right)}{{\tan}}^{{5}}{\left({u}\right)}{d}{u}$

Again apply the integral substitution: $\displaystyle{v}={\tan{{\left({u}\right)}}}$

$\displaystyle{d}{v}={{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({1}+{{v}}^{{2}}\right)}{{v}}^{{5}}{d}{v}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({{v}}^{{5}}+{{v}}^{{7}}\right)}{d}{v}$

apply the sum rule and power rule,

$\displaystyle=\frac{{1}}{{2}}{\left(\int{{v}}^{{5}}{d}{v}+\int{{v}}^{{7}}{d}{v}\right)}$

$=1/2{(v^(5+1)/(5+1))+(v^(7+1)/(7+1))}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{v}}^{{6}}}{{6}}+\frac{{{v}}^{{8}}}{{8}}\right)}$

substitute back $\displaystyle{v}={\tan{{\left({u}\right)}}}$ and $\displaystyle{u}={2}{x}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{{\tan}}^{{6}}{\left({2}{x}\right)}}}{{6}}+\frac{{{{\tan}}^{{8}}{\left({2}{x}\right)}}}{{8}}\right)}$

Add a constant C to the solution,

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{1}}{{6}}{{\tan}}^{{6}}{\left({2}{x}\right)}+\frac{{1}}{{8}}{{\tan}}^{{8}}{\left({2}{x}\right)}\right)}+{C}$

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$

Let's apply integral substitution: $\displaystyle{u}={2}{x}$

$\displaystyle{\left({d}{u}\right)}={2}{\left.{d}{x}\right.}$

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}=\int{{\tan}}^{{5}}{\left({u}\right)}{{\sec}}^{{4}}{\left({u}\right)}\frac{{{d}{u}}}{{2}}$

Take the constant out and rewrite the integral as,

$\displaystyle=\frac{{1}}{{2}}\int{{\sec}}^{{2}}{\left({u}\right)}{{\sec}}^{{2}}{\left({u}\right)}{{\tan}}^{{5}}{\left({u}\right)}{d}{u}$

Now use the trigonometric identity : $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({1}+{{\tan}}^{{2}}{\left({u}\right)}\right)}{{\sec}}^{{2}}{\left({u}\right)}{{\tan}}^{{5}}{\left({u}\right)}{d}{u}$

Again apply the integral substitution: $\displaystyle{v}={\tan{{\left({u}\right)}}}$

$\displaystyle{d}{v}={{\sec}}^{{2}}{\left({u}\right)}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({1}+{{v}}^{{2}}\right)}{{v}}^{{5}}{d}{v}$

$\displaystyle=\frac{{1}}{{2}}\int{\left({{v}}^{{5}}+{{v}}^{{7}}\right)}{d}{v}$

apply the sum rule and power rule,

$\displaystyle=\frac{{1}}{{2}}{\left(\int{{v}}^{{5}}{d}{v}+\int{{v}}^{{7}}{d}{v}\right)}$

$=1/2{(v^(5+1)/(5+1))+(v^(7+1)/(7+1))}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{v}}^{{6}}}{{6}}+\frac{{{v}}^{{8}}}{{8}}\right)}$

substitute back $\displaystyle{v}={\tan{{\left({u}\right)}}}$ and $\displaystyle{u}={2}{x}$

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{{{\tan}}^{{6}}{\left({2}{x}\right)}}}{{6}}+\frac{{{{\tan}}^{{8}}{\left({2}{x}\right)}}}{{8}}\right)}$

Add a constant C to the solution,

$\displaystyle=\frac{{1}}{{2}}{\left(\frac{{1}}{{6}}{{\tan}}^{{6}}{\left({2}{x}\right)}+\frac{{1}}{{8}}{{\tan}}^{{8}}{\left({2}{x}\right)}\right)}+{C}$

Notes

Wednesday, 12 March 2014

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Wednesday, 12 March 2014

Find the indefinite integral

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle\int{{\tan}}^{{5}}{\left({2}{x}\right)}{{\sec}}^{{4}}{\left({2}{x}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?