Fancy Integration Methods

• Jul 30th 2013, 06:19 PM
zachd77
Fancy Integration Methods
Out of curiosity, what are some integration methods that require special substitutions or unusual tactics that can be used to compute an integral? (Any method would do except commonly used methods like substitution, parts, partial fractions, and trig substitution.)
• Jul 30th 2013, 08:11 PM
Prove It
Re: Fancy Integration Methods
An interesting way to evaluate \displaystyle \begin{align*} \int{\arctan{(x)}\,dx} \end{align*} would be to consider trying to evaluate \displaystyle \begin{align*} \int{ \frac{2x}{1 + x^2}\,dx} \end{align*} in two different ways (just for convenience I'll leave out the integration constants).

Obviously, a substitution of the form \displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*} is appropriate, giving

\displaystyle \begin{align*} \int{ \frac{2x}{1 + x^2}\,dx} &= \int{ \frac{1}{u}\,du} \\ &= \ln{|u|} \\ &= \ln{ \left| 1 +x^2 \right|} \\ &= \ln{ \left( 1 + x^2 \right)} \textrm{ since } 1 + x^2 > 0 \textrm{ for all } x \end{align*}

Now let's suppose we didn't know that this substitution is appropriate, we could attempt using integration by parts, with \displaystyle \begin{align*} u = 2x \implies du = 2\,dx \end{align*} and \displaystyle \begin{align*} dv = \frac{1}{1 + x^2}\,dx \implies v = \arctan{(x)} \end{align*}, giving

\displaystyle \begin{align*} \int{ \frac{2x}{1 + x^2}\,dx} &= 2x\arctan{(x)} - \int{ 2\arctan{(x)}\,dx} \\ &= 2x\arctan{(x)} - 2\int{\arctan{(x)}\,dx} \end{align*}

So equating the two answers gives

\displaystyle \begin{align*} \ln{ \left( 1 + x^2 \right) } &= 2x\arctan{(x)} - 2\int{ \arctan{(x)}\,dx} \\ 2\int{ \arctan{(x)}\,dx} &= 2x\arctan{(x)} - \ln{ \left( 1 + x^2 \right)} \\ \int{ \arctan{(x)}\,dx} &= x\arctan{(x)} - \frac{1}{2}\ln{ \left( 1 + x^2 \right)} \end{align*}

and thus we can conclude

\displaystyle \begin{align*} \int{\arctan{(x)}\,dx} = x\arctan{(x)} - \frac{1}{2}\ln{ \left( 1 + x^2 \right)} + C \end{align*}
• Jul 31st 2013, 04:50 AM
zachd77
Re: Fancy Integration Methods
I know that that method works, but I want to find an even stranger method to compute an integral. Can someone elaborate onto the substitution u=x^(1/n) where n represents the least common multiple of two or more denominators of the powers of functions?