Blog: `int cos^3(theta) sin(theta) d theta, u = cos(theta)` Evaluate the integral by making the given substitution.

Wednesday, October 12, 2016

$\displaystyle\int{{\cos}}^{{3}}{\left(\theta\right)}{\sin{{\left(\theta\right)}}}{d}\theta,{u}={\cos{{\left(\theta\right)}}}$ Evaluate the integral by making the given substitution.

You need to evaluate the indefinite integral by performing the substitution $\displaystyle{u}={\cos{\theta}}$ , such that:

$\displaystyle{u}={\cos{\theta}}\Rightarrow{d}{u}=-{\sin{\theta}}\cdot{d}\theta\Rightarrow{\sin{\theta}}\cdot{d}\theta=-{d}{u}$

$\displaystyle\int{{\cos}}^{{3}}\theta\cdot{\sin{\theta}}\cdot{d}\theta=-\int{{u}}^{{3}}{d}{u}$

Using the formula $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{{u}}^{{{n}+{1}}}}}{{{n}+{1}}}+{c}$ yields

$\displaystyle-\int{{u}}^{{3}}{d}{u}=-\frac{{{{u}}^{{{3}+{1}}}}}{{{3}+{1}}}+{c}$

$\displaystyle-\int{{u}}^{{3}}{d}{u}=-\frac{{{{u}}^{{4}}}}{{4}}+{c}$

Replacing back $\displaystyle{\cos{\theta}}$ for $\displaystyle{u}$ yields:

$\displaystyle\int{{\cos}}^{{3}}\theta\cdot{\sin{\theta}}\cdot{d}\theta=-\frac{{{{\cos}}^{{4}}\theta}}{{4}}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{{\cos}}^{{3}}\theta\cdot{\sin{\theta}}\cdot{d}\theta=-\frac{{{{\cos}}^{{4}}\theta}}{{4}}+{c}.$

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Wednesday, October 12, 2016

$\displaystyle\int{{\cos}}^{{3}}{\left(\theta\right)}{\sin{{\left(\theta\right)}}}{d}\theta,{u}={\cos{{\left(\theta\right)}}}$ Evaluate the integral by making the given substitution.

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