Blog: `int arctan(4t) dt` Evaluate the integral

Sunday, January 30, 2011

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}{\left.{d}{t}\right.}$

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 integration by parts method,

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}{\left.{d}{t}\right.}={\arctan{{\left({4}{t}\right)}}}\cdot\int{1}{\left.{d}{t}\right.}-\int{\left(\frac{{d}}{{\left.{d}{t}}}{\left({\arctan{{\left({4}{t}\right)}}}\int{1}{\left.{d}{t}\right.}\right)}{\left.{d}{t}\right.}\right.}$

$\displaystyle={\arctan{{\left({4}{t}\right)}}}\cdot{t}-\int{\left(\frac{{4}}{{{{\left({4}{t}\right)}}^{{2}}+{1}}}\cdot{t}\right)}{\left.{d}{t}\right.}$

$\displaystyle={t}{\arctan{{\left({4}{t}\right)}}}-{4}\int\frac{{t}}{{{16}{{t}}^{{2}}+{1}}}{\left.{d}{t}\right.}$

Now let's evaluate $\displaystyle\int\frac{{t}}{{{16}{{t}}^{{2}}+{1}}}{\left.{d}{t}\right.}$ by using the method of substitution,

Substitute $\displaystyle{x}={16}{{t}}^{{2}}+{1},\Rightarrow{\left.{d}{x}\right.}={32}{t}{\left.{d}{t}\right.}$

$\displaystyle\int\frac{{t}}{{{16}{{t}}^{{2}}+{1}}}{\left.{d}{t}\right.}=\int\frac{{\left.{d}{x}}}{{{32}{x}}}$

$\displaystyle=\frac{{1}}{{32}}{\ln}{\left|{x}\right|}$

substitute back $\displaystyle{x}={16}{{t}}^{{2}}+{1}$

$\displaystyle=\frac{{1}}{{32}}{\ln}{\left|{16}{{t}}^{{2}}+{1}\right|}$

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}={t}\cdot{\arctan{{\left({4}{t}\right)}}}-\frac{{4}}{{32}}{\ln}{\left|{16}{{t}}^{{2}}+{1}\right|}+{C}$

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}={t}\cdot{\arctan{{\left({4}{t}\right)}}}-\frac{{1}}{{8}}{\ln}{\left|{16}{{t}}^{{2}}+{1}\right|}+{C}$

C is a constant.

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Sunday, January 30, 2011

$\displaystyle\int{\arctan{{\left({4}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

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