# Thread: Integration With Inverse Trig Function

1. ## Integration With Inverse Trig Function

I want you to look at question (b) in the picture provided and tell me where (1/3) outside the integral came from. Is there factoring involved

I am to use the formula below.

Let INT = integral

INT (du)/(a^2+u^2) = (1/a)*arctan(u/a) + C

2. ## Re: Integration With Inverse Trig Function

Originally Posted by nycmath
I want you to look at question (b) in the picture provided and tell me where (1/3) outside the integral came from. Is there factoring involved

I am to use the formula below.

Let INT = integral

INT (du)/(a^2+u^2) = (1/a)*arctan(u/a) + C
it balances the 3 they added to the numerator next to the dx

3. ## Re: Integration With Inverse Trig Function

It's because when the substitution \displaystyle \begin{align*} u = 3x \end{align*} is applied, the differential is \displaystyle \begin{align*} du = 3\,dx \end{align*}. To get the extra factor of 3 on the inside, a factor of 1/3 has to be multiplied outside.

4. ## Re: Integration With Inverse Trig Function

The textbook has plenty of questions at the end of the chapter. Can someone completely work out question (b) with steps to guide me through the rest?

5. ## Re: Integration With Inverse Trig Function

It HAS been completely worked out for you! Replace the 3x with u, replace the 3dx with du, and then perform the integration using the arctan integral rule you already know.

6. ## Re: Integration With Inverse Trig Function

Originally Posted by nycmath
The textbook has plenty of questions at the end of the chapter. Can someone completely work out question (b) with steps to guide me through the rest?
$\displaystyle{\int}\dfrac{dx}{2+9x^2}$

and you want to use the template

$\displaystyle{\int}\dfrac{dx}{x^2+a^2}=\dfrac{1}{ a}\arctan\left(\frac{x}{a}\right)+C$

\large \begin{align*} &\displaystyle{\int}\dfrac{dx}{2+9x^2}= \\ \\ &\frac{1}{9}\displaystyle{\int}\dfrac{dx}{\frac{2} {9}+x^2} =\\ \\ &\frac{1}{9}\displaystyle{\int}\dfrac{dx}{ \left(\sqrt{ \frac{2}{9}}\right)^2+x^2} \\ \\ &\mbox{we see that }a=\sqrt{\frac{2}{9}} \\ \\ &\frac{1}{9}\displaystyle{\int}\dfrac{dx}{ \left(\sqrt{ \frac{2}{9}}\right)^2+x^2}= \\ \\ &\frac{1}{9} \sqrt{\dfrac{9}{2}}\arctan\left(\sqrt{ \frac{9}{2}}x\right)+C = \\ \\ &\frac{1}{3\sqrt{2}} \arctan\left(\frac{3x}{\sqrt{2}}\right)+C \end{align*}

7. ## Re: Integration With Inverse Trig Function

Thank you, romsek.