The Notes: `int_0^(pi/4) sqrt(1 - cos(4 theta)) d theta` Evaluate the integral

Thursday, April 29, 2010

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}\sqrt{{{1}-{\cos{{\left({4}\theta\right)}}}}}{d}\theta$

Let' first compute the indefinite integral,

$\displaystyle\int\sqrt{{{1}-{\cos{{\left({4}\theta\right)}}}}}{d}\theta$

Apply the integral substitution.

Let $\displaystyle\theta=\frac{{u}}{{2}}$

$\displaystyle\Rightarrow{d}\theta=\frac{{1}}{{2}}{d}{u}$

$\displaystyle=\int\frac{{1}}{{2}}\sqrt{{{1}-{\cos{{2}}}{u}}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}\int\sqrt{{{1}-{\cos{{2}}}{u}}}{d}{u}$

use the identity: $\displaystyle{\cos{{\left({2}{x}\right)}}}={1}-{2}{{\sin}}^{{2}}{\left({x}\right)}$

$\displaystyle=\frac{{1}}{{2}}\int\sqrt{{{1}-{\left({1}-{2}{{\sin}}^{{2}}{u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}\int\sqrt{{{2}{{\sin}}^{{2}}{u}}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}\int\sqrt{{{2}}}\sqrt{{{{\sin}}^{{2}}{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{\sqrt{{{2}}}}{{2}}\int\sqrt{{{{\sin}}^{{2}}{\left({u}\right)}}}{d}{u}$

$\displaystyle={\left(\frac{{1}}{\sqrt{{{2}}}}\right)}\int{\sin{{\left({u}\right)}}}{d}{u}$

Change the bounds from 0 to pi/4 to lower bound 0 and upper bound pi/2.

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\sin{{\left({u}\right)}}}{d}{u}$

$\displaystyle-\frac{{\cos{{\left({u}\right)}}}}{\sqrt{{{2}}}}{\)}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}$

$\displaystyle={0}--\frac{{1}}{\sqrt{{{2}}}}=\frac{{1}}{\sqrt{{{2}}}}$

The Notes