`int_0^(pi/6) sqrt(1 + cos(2x)) dx` Evaluate the integral

Friday, December 30, 2016

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{\cos{{\left({2}{x}\right)}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{\cos{{\left({2}{x}\right)}}}}}{\left.{d}{x}\right.}$

Rewrite the integrand by using the identity: $\displaystyle{\cos{{\left({2}{x}\right)}}}={2}{{\cos}}^{{2}}{\left({x}\right)}-{1}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{1}+{2}{{\cos}}^{{2}}{\left({x}\right)}-{1}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{6}}}}}\sqrt{{{2}{{\cos}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle=\sqrt{{{2}}}{\int_{{0}}^{{\frac{\pi}{{6}}}}}{\left|{\cos{{\left({x}\right)}}}\right|}{\left.{d}{x}\right.}$

$\displaystyle=\sqrt{{{2}}}{\int_{{0}}^{{\frac{\pi}{{6}}}}}{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}$ (as $\displaystyle{0}\le{x}\le\frac{\pi}{{6}}\Rightarrow{\cos{{\left({x}\right)}}}\ge{0}\Rightarrow{\left|{\cos{{\left({x}\right)}}}\right|}={\cos{{\left({x}\right)}}}$ )

$\displaystyle=\sqrt{{{2}}}{{\left[{\sin{{\left({x}\right)}}}\right]}_{{0}}^{{\frac{\pi}{{6}}}}}$

$\displaystyle=\sqrt{{{2}}}{\left[{\sin{{\left(\frac{\pi}{{6}}\right)}}}-{\sin{{\left({0}\right)}}}\right]}$

$\displaystyle=\sqrt{{{2}}}{\left[\frac{{1}}{{2}}-{0}\right]}$

$\displaystyle=\frac{\sqrt{{{2}}}}{{2}}$

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