Blog: `int (dx)/(sqrt(1 - x^2)sin^(-1)(x))` Evaluate the indefinite integral.

Saturday, June 15, 2013

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{1}-{{x}}^{{2}}}}{{\sin}}^{{-{1}}}{\left({x}\right)}}}$ Evaluate the indefinite integral.

You need to evaluate the indefinite integral, using the following substitution, such that:

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

Replacing t for x yields:

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

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{{t}}={\ln}{\left|{t}\right|}+{c}$

Replacing back $\displaystyle{{\sin}}^{{-{1}}}{x}$ for t yields:

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

Hence, evaluating the indefinite integral, using the substitution $\displaystyle{{\sin}}^{{-{1}}}{x}={t}$ , yields $\displaystyle\int\frac{{\left.{d}{x}}}{{{\left({{\sin}}^{{-{1}}}{x}\right)}\sqrt{{{1}-{{x}}^{{2}}}}}}={\ln}{\left|{{\sin}}^{{-{1}}}{x}\right|}+{c}.$

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Saturday, June 15, 2013

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{1}-{{x}}^{{2}}}}{{\sin}}^{{-{1}}}{\left({x}\right)}}}$ Evaluate the indefinite integral.

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