`int (2 - tan(theta/4)) d theta` Find the indefinite integral.

Tuesday, January 12, 2016

$\displaystyle\int{\left({2}-{\tan{{\left(\frac{\theta}{{4}}\right)}}}\right)}{d}\theta$ Find the indefinite integral.

$\displaystyle\int{\left({2}-{\tan{{\left(\frac{\theta}{{4}}\right)}}}\right)}{d}\theta=$

Use additivity of integral: $\displaystyle\int{\left({f{{\left({x}\right)}}}\pm{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\pm\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}.$ $\displaystyle\int{2}{d}\theta-\int{\tan{{\left(\frac{\theta}{{4}}\right)}}}{d}\theta=$

Since the first integral is easy $\displaystyle\int{2}{d}\theta={2}\theta+{C}$ we will concentrate on the second integral. To solve it we will make substitution: $\displaystyle{u}=\frac{\theta}{{4}},$ $\displaystyle{d}{u}=\frac{{{d}\theta}}{{4}}\Rightarrow{d}\theta={4}{d}{u}$

$\displaystyle\int{\tan{{\left(\frac{\theta}{{4}}\right)}}}{d}\theta={4}\int{\tan{{u}}}{d}{u}=-{4}{\ln}{\left|{\cos{{u}}}\right|}+{C}$

Return the substitution.

$\displaystyle-{4}{\ln}{\left|{\cos{{\left(\frac{\theta}{{4}}\right)}}}\right|}+{C}$

Now we subtract the two integrals to obtain the final result.

$\displaystyle{2}\theta-{\left(-{4}{\ln}{\left|{\cos{{\left(\frac{\theta}{{4}}\right)}}}\right|}\right)}+{C}={2}\theta+{4}{\ln}{\left|{\cos{{\left(\frac{\theta}{{4}}\right)}}}\right|}+{C}$

Ninenotes

Tuesday, January 12, 2016

$\displaystyle\int{\left({2}-{\tan{{\left(\frac{\theta}{{4}}\right)}}}\right)}{d}\theta$ Find the indefinite integral.

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