- #1
Prove It
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Hi everyone. Just for fun I thought we could post some of the more interesting ways we know of to evaluate integrals :)
For starters, to evaluate [math]\displaystyle \begin{align*} \int{\arctan{(x)}\,dx} \end{align*}[/math], first we consider the integral [math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} \end{align*}[/math]. For simplicity, we'll leave out integration constants til the end...
We can integrate this using a substitution [math]\displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*}[/math] and the integral becomes
[math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{\frac{1}{u}\,du} \\ &= \ln{|u|} + C \\ &= \ln{ \left| 1 + x^2 \right| } + C \\ &= \ln{ \left( 1 + x^2 \right) } \textrm{ since } 1 + x^2 > 0 \textrm{ for all } x \in \mathbf{R} \end{align*}[/math]
Now supposing we wanted to evaluate the integral in a different way, using integration by parts with [math]\displaystyle \begin{align*} u = 2x \implies du = 2\,dx \end{align*}[/math] and [math]\displaystyle \begin{align*} dv = \frac{1}{1 + x^2}\,dx \implies v = \arctan{(x)} \end{align*}[/math], then we would have
[math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{2x \left( \frac{1}{1 + x^2} \right) dx} \\ &= 2x\arctan{(x)} - \int{2\arctan{(x)}\,dx} \\ &= 2x\arctan{(x)} - 2\int{\arctan{(x)}\,dx} \end{align*}[/math]
Now equating these gives
[math]\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) } + C \end{align*}[/math]
Q.E.D.
For starters, to evaluate [math]\displaystyle \begin{align*} \int{\arctan{(x)}\,dx} \end{align*}[/math], first we consider the integral [math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} \end{align*}[/math]. For simplicity, we'll leave out integration constants til the end...
We can integrate this using a substitution [math]\displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*}[/math] and the integral becomes
[math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{\frac{1}{u}\,du} \\ &= \ln{|u|} + C \\ &= \ln{ \left| 1 + x^2 \right| } + C \\ &= \ln{ \left( 1 + x^2 \right) } \textrm{ since } 1 + x^2 > 0 \textrm{ for all } x \in \mathbf{R} \end{align*}[/math]
Now supposing we wanted to evaluate the integral in a different way, using integration by parts with [math]\displaystyle \begin{align*} u = 2x \implies du = 2\,dx \end{align*}[/math] and [math]\displaystyle \begin{align*} dv = \frac{1}{1 + x^2}\,dx \implies v = \arctan{(x)} \end{align*}[/math], then we would have
[math]\displaystyle \begin{align*} \int{\frac{2x}{1 + x^2}\,dx} &= \int{2x \left( \frac{1}{1 + x^2} \right) dx} \\ &= 2x\arctan{(x)} - \int{2\arctan{(x)}\,dx} \\ &= 2x\arctan{(x)} - 2\int{\arctan{(x)}\,dx} \end{align*}[/math]
Now equating these gives
[math]\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) } + C \end{align*}[/math]
Q.E.D.