Blog: `int_0^(1/2) cos^(-1) x dx` Evaluate the integral

Saturday, December 14, 2013

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}{{\cos}}^{{-{1}}}{x}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}{{\cos}}^{{-{1}}}{x}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral by using the method of integration by parts,

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

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

$\displaystyle={x}{{\cos}}^{{-{1}}}{x}+\int\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Now let's evaluate $\displaystyle\int\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}$ using the method of substitution,

Let's substitute $\displaystyle{t}={1}-{{x}}^{{2}}$

$\displaystyle\Rightarrow{\left.{d}{t}\right.}=-{2}{x}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}{\left.{d}{x}\right.}=\int\frac{{\left.{d}{t}}}{{-{2}\sqrt{{{t}}}}}$

$\displaystyle=-\frac{{1}}{{2}}\int\frac{{\left.{d}{t}}}{\sqrt{{{t}}}}$

$\displaystyle=-\frac{{1}}{{2}}{\left(\frac{{{t}}^{{-\frac{{1}}{{2}}+{1}}}}{{-\frac{{1}}{{2}}+{1}}}\right)}$

$\displaystyle=-\frac{{1}}{{2}}{\left(\frac{{{t}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}\right)}$

$\displaystyle=-{{t}}^{{\frac{{1}}{{2}}}}$

substitute back $\displaystyle{t}={1}-{{x}}^{{2}}$

$\displaystyle=-\sqrt{{{1}-{{x}}^{{2}}}}$

$\displaystyle\therefore\int{{\cos}}^{{-{1}}}{x}{\left.{d}{x}\right.}={x}{{\cos}}^{{-{1}}}{x}-\sqrt{{{1}-{{x}}^{{2}}}}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{\frac{{1}}{{2}}}}}{{\cos}}^{{-{1}}}{x}={{\left[{x}{{\cos}}^{{-{1}}}{x}-\sqrt{{{1}-{{x}}^{{2}}}}\right]}_{{0}}^{{\frac{{1}}{{2}}}}}$

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