`int sin^2(pi x) cos^5(pi x) dx` Evaluate the integral

Friday, February 6, 2015

$\displaystyle\int{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{5}}{\left(\pi{x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{5}}{\left(\pi{x}\right)}{\left.{d}{x}\right.}$

To solve, apply the Pythagorean identity $\displaystyle{{\sin}}^{{2}}\theta+{{\cos}}^{{2}}\theta={1}$ repeatedly until the integral is in the form $\displaystyle\int{{u}}^{{n}}{d}{u}$ .

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

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

$\displaystyle=\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{3}}{\left(\pi{x}\right)}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{{\cos}}^{{3}}{\left(\pi{x}\right)}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{{\cos}}^{{2}}{\left(\pi{x}\right)}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{{\cos}}^{{2}}{\left(\pi{x}\right)}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{\left({1}-{{\sin}}^{{2}}{\left(\pi{x}\right)}\right)}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{\left({1}-{{\sin}}^{{2}}{\left(\pi{x}\right)}\right)}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}+{{\sin}}^{{6}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}-{2}{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}+{{\sin}}^{{6}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\int{{\sin}}^{{2}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}-\int{2}{{\sin}}^{{4}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}+\int{{\sin}}^{{6}}{\left(\pi{x}\right)}{\cos{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}$

To take the integral of this, apply u-substitution method.

$\displaystyle{u}={\sin{{\left(\pi{x}\right)}}}$

...

$\displaystyle\int{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{5}}{\left(\pi{x}\right)}{\left.{d}{x}\right.}$

To solve, apply the Pythagorean identity $\displaystyle{{\sin}}^{{2}}\theta+{{\cos}}^{{2}}\theta={1}$ repeatedly until the integral is in the form $\displaystyle\int{{u}}^{{n}}{d}{u}$ .

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

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

$\displaystyle=\int{\left[{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{3}}{\left(\pi{x}\right)}-{{\sin}}^{{4}}{\left(\pi{x}\right)}{{\cos}}^{{3}}{\left(\pi{x}\right)}\right]}{\left.{d}{x}\right.}$

To take the integral of this, apply u-substitution method.

$\displaystyle{u}={\sin{{\left(\pi{x}\right)}}}$

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

$\displaystyle\frac{{{d}{u}}}{\pi}={\cos{{\left(\pi{x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{{u}}^{{2}}\cdot\frac{{{d}{u}}}{\pi}-\int{2}{{u}}^{{4}}\cdot\frac{{{d}{u}}}{\pi}+\int{{u}}^{{6}}\cdot\frac{{{d}{u}}}{\pi}$

$\displaystyle=\frac{{1}}{\pi}\int{{u}}^{{2}}{d}{u}-\frac{{2}}{\pi}\int{{u}}^{{4}}{d}{u}+\int\frac{{1}}{\pi}{{u}}^{{6}}{d}{u}$

$\displaystyle=\frac{{1}}{\pi}\cdot\frac{{{u}}^{{2}}}{{3}}-\frac{{2}}{\pi}\cdot\frac{{{u}}^{{5}}}{{5}}+\frac{{1}}{\pi}\cdot\frac{{{u}}^{{7}}}{{7}}+{C}$

$\displaystyle=\frac{{{u}}^{{2}}}{{{3}\pi}}-\frac{{{2}{{u}}^{{5}}}}{{{5}\pi}}+\frac{{{u}}^{{7}}}{{{7}\pi}}+{C}$

And, substitute back $\displaystyle{u}={\sin{{\left(\pi{x}\right)}}}$ .

$\displaystyle=\frac{{{{\sin}}^{{2}}{\left(\pi{x}\right)}}}{{{3}\pi}}-\frac{{{2}{{\sin}}^{{5}}{\left(\pi{x}\right)}}}{{{5}\pi}}+\frac{{{{\sin}}^{{7}}{\left(\pi{x}\right)}}}{{{7}\pi}}+{C}$

Therefore, $\displaystyle\int{{\sin}}^{{2}}{\left(\pi{x}\right)}{{\cos}}^{{5}}{\left(\pi{x}\right)}{\left.{d}{x}\right.}=\frac{{{{\sin}}^{{2}}{\left(\pi{x}\right)}}}{{{3}\pi}}-\frac{{{2}{{\sin}}^{{5}}{\left(\pi{x}\right)}}}{{{5}\pi}}+\frac{{{{\sin}}^{{7}}{\left(\pi{x}\right)}}}{{{7}\pi}}+{C}.$