Blog: `int sqrt(cot(x)) csc^2(x) dx` Evaluate the indefinite integral.

Monday, October 24, 2011

$\displaystyle\int\sqrt{{{\cot{{\left({x}\right)}}}}}{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

You need to use the following substitution $\displaystyle{\cot{{x}}}={t}$ , such that:

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

$\displaystyle\int\sqrt{{{\cot{{x}}}}}\cdot{{\csc}}^{{2}}{x}{\left.{d}{x}\right.}=-\int\sqrt{{t}}{\left.{d}{t}\right.}$

$\displaystyle-\int\sqrt{{t}}{\left.{d}{t}\right.}=-\frac{{{{t}}^{{\frac{{3}}{{2}}}}}}{{\frac{{3}}{{2}}}}+{c}$

Replacing back cot x for t yields:

$\displaystyle\int\sqrt{{{\cot{{x}}}}}\cdot{{\csc}}^{{2}}{x}{\left.{d}{x}\right.}=-{\left(\frac{{2}}{{3}}\right)}{\left({{\left({\cot{{x}}}\right)}}^{{\frac{{3}}{{2}}}}\right)}+{c}$

Hence, evaluating the indefinite integral, yields $\displaystyle\int\sqrt{{{\cot{{x}}}}}\cdot{{\csc}}^{{2}}{x}{\left.{d}{x}\right.}=-{\left(\frac{{2}}{{3}}\right)}{\left({{\left({\cot{{x}}}\right)}}^{{\frac{{3}}{{2}}}}\right)}+{c}$

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Monday, October 24, 2011

$\displaystyle\int\sqrt{{{\cot{{\left({x}\right)}}}}}{{\csc}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

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