The Notes: `int sin^2(x) cos^3(x) dx` Evaluate the integral

Tuesday, November 5, 2013

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

Use Pythagorean trigonometric identity: $\displaystyle{{\sin}}^{{2}}{x}+{{\cos}}^{{2}}{x}={1}\Rightarrow{{\cos}}^{{2}}{x}={1}-{{\sin}}^{{2}}{x}$

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

$\displaystyle\int{{\sin}}^{{2}}{x}{\cos{{x}}}{\left.{d}{x}\right.}-\int{{\sin}}^{{4}}{x}{\cos{{x}}}{\left.{d}{x}\right.}$

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

$\displaystyle\int{{t}}^{{2}}{\left.{d}{t}\right.}-\int{{t}}^{{4}}{\left.{d}{t}\right.}=\frac{{{t}}^{{3}}}{{3}}-\frac{{{t}}^{{5}}}{{5}}+{C}=$

Return substitution to get the solution.

$\displaystyle\frac{{{{\sin}}^{{3}}{x}}}{{3}}-\frac{{{{\sin}}^{{5}}{x}}}{{5}}+{C}$

The Notes