The Notes: `int x tan^2(x) dx` Evaluate the integral

Friday, December 5, 2008

$\displaystyle\int{x}{{\tan}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{x}{{\tan}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

Let's evaluate the above integral by using the method of integration by parts,

If f(x) and g(x) are differentiable functions , then

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

If we write f(x)=u and g'(x)=v, then

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

Let's denote u=x and $\displaystyle{v}={{\tan}}^{{2}}{\left({x}\right)}$

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

Now let's use the identity: $\displaystyle{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}-{1}$

$\displaystyle={x}\cdot\int{\left({{\sec}}^{{2}}{\left({x}\right)}-{1}\right)}{\left.{d}{x}\right.}-\int{\left(\int{\left({{\sec}}^{{2}}{\left({x}\right)}-{1}\right)}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}\cdot{\left(\int{{\sec}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}-\int{1}{\left.{d}{x}\right.}\right)}-\int{\left(\int{\left({{\sec}}^{{2}}{\left({x}\right)}-{1}\right)}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}\cdot{\left({\tan{{\left({x}\right)}}}-{x}\right)}-\int{\left({\tan{{\left({x}\right)}}}-{x}\right)}{\left.{d}{x}\right.}$

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

$\displaystyle={x}{\tan{{\left({x}\right)}}}-{{x}}^{{2}}-{\left(-{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}\right)}+\frac{{{x}}^{{2}}}{{2}}$

$\displaystyle={x}{\tan{{\left({x}\right)}}}-{{x}}^{{2}}+\frac{{{x}}^{{2}}}{{2}}+{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}$

$\displaystyle={x}{\tan{{\left({x}\right)}}}+{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}-\frac{{{x}}^{{2}}}{{2}}$

Add a constant C to the solution,

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

The Notes

Friday, December 5, 2008

$\displaystyle\int{x}{{\tan}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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