# Thread: Integrals Involving Inverse Trig Functions

1. ## Integrals Involving Inverse Trig Functions

Now for integrals involving inverse trig functions.

A couple problems:

$\int \frac {1}{1 + 4x^2}dx$

For this one, I know that $\frac {d}{dx} \arctan {x} = \frac {1}{1-x^2}$ but I don't know what to do with the 4 in there.

$\int \frac {1}{x^2 - 6x + 13}dx$
This one I have no clue.

Thanks

2. Originally Posted by Chicken Enchilada
Now for integrals involving inverse trig functions.

A couple problems:

$\int \frac {1}{1 + 4x^2}dx$

For this one, I know that $\frac {d}{dx} \arctan {x} = \frac {1}{1-x^2}$ but I don't know what to do with the 4 in there.

$\int \frac {1}{x^2 - 6x + 13}dx$
This one I have no clue.

Thanks
For the first one, note that $\int\frac{\,dx}{4x^2+1}=\int\frac{\,dx}{\left(2x\r ight)^2+1}$. Then let $u=2x$.

For the second one, note that $\int\frac{\,dx}{x^2-6x+13}=\int\frac{\,dx}{(x^2-6x+9)+4}=\int\frac{\,dx}{(x-3)^2+4}=\tfrac{1}{4}\int\frac{\,dx}{\left(\frac{x-3}{2}\right)^2+1}$. Then let $u=\frac{x-3}{2}$.

Can you take it from here?

3. Originally Posted by Chris L T521
For the first one, note that $\int\frac{\,dx}{4x^2+1}=\int\frac{\,dx}{\left(2x\r ight)^2+1}$. Then let $u=2x$.

For the second one, note that $\int\frac{\,dx}{x^2-6x+13}=\int\frac{\,dx}{(x^2-6x+9)+4}=\int\frac{\,dx}{(x-3)^2+4}=\tfrac{1}{4}\int\frac{\,dx}{\left(\frac{x-3}{2}\right)^2+1}$. Then let $u=\frac{x-3}{2}$.

Can you take it from here?
So for the first one the answer is:
$\frac {1}{2} \arctan (2x) + c$

and for the second:
$\frac {1}{2} \arctan (\frac {x-3}{2}) + c$

Sweet! That was actually very simple. You taught me how to solve these problems in a matter of seconds, whereas my prof couldn't do it in 2 hours. It really helps if you can see the problem being solved unambiguously with every little step shown. I think my prof tends to skip a bunch, and everything comes out muddy to me.

I appreciate the help Chris. Math Help Forum rules!