`int t sec^2 2t dt` Evaluate the integral

Tuesday, May 26, 2015

$\displaystyle\int{t}{{\sec}}^{{2}}{2}{t}{\left.{d}{t}\right.}$ Evaluate the integral

$\displaystyle\int{t}{{\sec}}^{{2}}{\left({2}{t}\right)}{\left.{d}{t}\right.}$

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

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Now using
the above method of integration by parts,

$\displaystyle\int{t}{{\sec}}^{{2}}{\left({2}{t}\right)}{\left.{d}{t}\right.}={t}\int{{\sec}}^{{2}}{\left({2}{t}\right)}{\left.{d}{t}\right.}-\int{\left(\frac{{d}}{{\left.{d}{t}}}{\left({t}\right)}\int{{\sec{{t}}}}^{{2}}{\left({2}{t}\right)}\right)}{\left.{d}{t}\right.}$

$\displaystyle={t}\cdot\frac{{\tan{{\left({2}{t}\right)}}}}{{2}}-\int{\left({1}\cdot\frac{{\tan{{\left({2}{t}\right)}}}}{{2}}\right)}{\left.{d}{t}\right.}$

$\displaystyle=\frac{{1}}{{2}}{t}\cdot{\tan{{\left({2}{t}\right)}}}-\frac{{1}}{{2}}\int{\tan{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$

Now let's evaluate $\displaystyle\int{\tan{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ by
using the method of substitution,

Substitute $\displaystyle{x}={\cos{{\left({2}{t}\right)}}},\Rightarrow{\left.{d}{x}\right.}=-{2}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle\int{\tan{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=\int{\left(\frac{{\sin{{\left({2}{t}\right)}}}}{{\cos{{\left({2}{t}\right)}}}}\right)}{\left.{d}{t}\right.}$

$\displaystyle=\int\frac{{\left.{d}{x}}}{{-{2}{x}}}$

$\displaystyle=-\frac{{1}}{{2}}{\ln}{\left|{x}\right|}$

substitute back $\displaystyle{x}={\cos{{\left({2}{t}\right)}}}$

$\displaystyle=-\frac{{1}}{{2}}{\ln}{\left|{\cos{{\left({2}{t}\right)}}}\right|}$

$\displaystyle\int{t}{{\sec}}^{{2}}{\left({2}{t}\right)}{\left.{d}{t}\right.}=\frac{{1}}{{2}}{t}\cdot{\tan{{\left({2}{t}\right)}}}-\frac{{1}}{{2}}{\left(-\frac{{1}}{{2}}{\ln}{\left|{\cos{{\left({2}{t}\right)}}}\right|}+{C}\right.}$

C is a
constant

$\displaystyle\int{t}{{\sec}}^{{2}}{\left({2}{t}\right)}{\left.{d}{t}\right.}=\frac{{1}}{{2}}{t}\cdot{\tan{{\left({2}{t}\right)}}}+\frac{{1}}{{4}}{\ln}{\left|{\cos{{\left({2}{t}\right)}}}\right|}+{C}$

Spooky Notes

Tuesday, May 26, 2015

$\displaystyle\int{t}{{\sec}}^{{2}}{2}{t}{\left.{d}{t}\right.}$ Evaluate the integral

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