The Notes: `int x sin^3(x) dx` Evaluate the integral

Wednesday, April 10, 2013

$\displaystyle\int{x}{{\sin}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

You need to use integration by parts, such that:

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

$\displaystyle{u}={x}\Rightarrow{d}{u}={\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{\sin}}^{{3}}{x}\Rightarrow{v}=\int{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}$

You need to solve the integral $\displaystyle\int{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}$ such that:

$\displaystyle\int{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}=\int{{\sin}}^{{2}}{x}\cdot{\sin{{x}}}{\left.{d}{x}\right.}$

Replace 1 $\displaystyle-{{\cos}}^{{2}}{x}$ for $\displaystyle{{\sin}}^{{2}}{x}$ , such that:

$\displaystyle\int{{\sin}}^{{2}}{x}\cdot{\sin{{x}}}{\left.{d}{x}\right.}=\int{\left({1}-{{\cos}}^{{2}}{x}\right)}\cdot{\sin{{x}}}{\left.{d}{x}\right.}$

Use substitution $\displaystyle{\cos{{x}}}={t}\Rightarrow-{\sin{{x}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}:$

$\displaystyle\int{\left({1}-{{\cos}}^{{2}}{x}\right)}\cdot{\sin{{x}}}{\left.{d}{x}\right.}=\int{\left({1}-{{t}}^{{2}}\right)}\cdot{\left(-{\left.{d}{t}\right.}\right)}$

$\displaystyle\int{\left({1}-{{t}}^{{2}}\right)}\cdot{\left(-{\left.{d}{t}\right.}\right)}=\int{\left({{t}}^{{2}}-{1}\right)}{\left.{d}{t}\right.}$

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

$\displaystyle\int{\left({{t}}^{{2}}-{1}\right)}{\left.{d}{t}\right.}=\frac{{{t}}^{{3}}}{{3}}-{t}+{c}$

Replace back cos x for t:

$\displaystyle\int{\left({1}-{{\cos}}^{{2}}{x}\right)}\cdot{\sin{{x}}}{\left.{d}{x}\right.}=\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}+{c}$

Hence, $\displaystyle{v}=\int{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}=\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}$

Using parts, yields:

$\displaystyle\int{x}{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}={x}\cdot{\left(\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}\right)}-\int{\left(\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle\int{x}{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}={x}\cdot{\left(\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}\right)}-\int\frac{{{{\cos}}^{{3}}{x}}}{{3}}{\left.{d}{x}\right.}+\int{\cos{{x}}}{\left.{d}{x}\right.}$

You need to solve the integral $\displaystyle\int\frac{{{{\cos}}^{{3}}{x}}}{{3}}{\left.{d}{x}\right.}$ , such that:

$\displaystyle\int\frac{{{{\cos}}^{{3}}{x}}}{{3}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{3}}\right)}\int{{\cos}}^{{2}}{x}\cdot{\cos{{x}}}{\left.{d}{x}\right.}$

Replace $\displaystyle{1}-{{\sin}}^{{2}}{x}$ for $\displaystyle{{\cos}}^{{2}}{x}:$

$\displaystyle{\left(\frac{{1}}{{3}}\right)}\int{\left({1}-{{\sin}}^{{2}}{x}\right)}\cdot{\cos{{x}}}{\left.{d}{x}\right.}$

Use substitution $\displaystyle{\sin{{x}}}={t}\Rightarrow{\cos{{x}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}$

$\displaystyle{\left(\frac{{1}}{{3}}\right)}\int{\left({1}-{{\sin}}^{{2}}{x}\right)}\cdot{\cos{{x}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{3}}\right)}\cdot\int{\left({1}-{{t}}^{{2}}\right)}{\left.{d}{t}\right.}={\left(\frac{{1}}{{3}}\right)}{t}-{\left(\frac{{1}}{{3}}\right)}\frac{{{{t}}^{{3}}}}{{3}}+{c}$

$\displaystyle{\left(\frac{{1}}{{3}}\right)}\int{\left({1}-{{\sin}}^{{2}}{x}\right)}\cdot{\cos{{x}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{3}}\right)}{\sin{{x}}}-{\left(\frac{{1}}{{3}}\right)}\frac{{{{\sin}}^{{3}}{x}}}{{3}}+{c}$

Hence, the result of integration is: $\displaystyle\int{x}{{\sin}}^{{3}}{x}{\left.{d}{x}\right.}={x}\cdot{\left(\frac{{{{\cos}}^{{3}}{x}}}{{3}}-{\cos{{x}}}\right)}-{\left({\left(\frac{{1}}{{3}}\right)}{\sin{{x}}}-{\left(\frac{{1}}{{3}}\right)}\frac{{{{\sin}}^{{3}}{x}}}{{3}}\right)}+{\sin{{x}}}+{c}$

The Notes

Wednesday, April 10, 2013

$\displaystyle\int{x}{{\sin}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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