Blog: `int sin^-1 x dx` Evaluate the integral

Monday, April 18, 2011

$\displaystyle\int{{\sin}}^{{-{{1}}}}{x}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{{\sin}}^{{-{{1}}}}{x}{\left.{d}{x}\right.}$

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we write f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

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

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

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

Now evaluate using the method of substitution,

Substitute $\displaystyle{t}={1}-{{x}}^{{2}},\Rightarrow{\left.{d}{t}\right.}=-{2}{x}{\left.{d}{x}\right.}$

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

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

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

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

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

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

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

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

adding constant C to the solution,

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

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Monday, April 18, 2011

$\displaystyle\int{{\sin}}^{{-{{1}}}}{x}{\left.{d}{x}\right.}$ Evaluate the integral

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