`int (sec(2x) + tan(2x)) dx` Find the indefinite integral.

Sunday, July 3, 2011

$\displaystyle\int{\left({\sec{{\left({2}{x}\right)}}}+{\tan{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral.

$\displaystyle\int{\left({\sec{{\left({2}{x}\right)}}}+{\tan{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}=$

Use additivity of integral: $\displaystyle\int{\left({f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}+\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}.$ $\displaystyle\int{\sec{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}+\int{\tan{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}=$

Make the same substitution for both integrals: $\displaystyle{u}={2}{x},$ $\displaystyle{d}{u}={2}{\left.{d}{x}\right.}\Rightarrow{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{2}}$

$\displaystyle\frac{{1}}{{2}}\int{\sec{{u}}}{d}{u}+\frac{{1}}{{2}}\int{\tan{{u}}}{d}{u}=$

Now we have table integrals.

$\displaystyle\frac{{1}}{{2}}{\ln}{\left|{\sec{{u}}}+{\tan{{u}}}\right|}-\frac{{1}}{{2}}{\ln}{\left|{\cos{{u}}}\right|}+{C}$

Return the substitution to obtain the final result.

$\displaystyle\frac{{1}}{{2}}{\ln}{\left|{\sec{{\left({2}{x}\right)}}}+{\tan{{\left({2}{x}\right)}}}\right|}-\frac{{1}}{{2}}{\ln}{\left|{\cos{{\left({2}{x}\right)}}}\right|}+{C}$

Ninenotes

Sunday, July 3, 2011

$\displaystyle\int{\left({\sec{{\left({2}{x}\right)}}}+{\tan{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral.

No comments:

Post a Comment

find square roots of -1+2i