1. ## Trig Substitution

I need to integrate

$\int 3/ (4-3x^2)^{1/2}$

I know I'm supposed to use the substitution x= a sin theta but I don't know what happens to the coefficient 3...

2. Originally Posted by Hellreaver
I need to integrate

$\int 3/ (4-3x^2)^{1/2}$

I know I'm supposed to use the substitution x= a sin theta but I don't know what happens to the coefficient 3...
Try using the substitution $\sqrt{3}x=2\sin\theta$. If you get stuck, post back.

3. Hello, Hellreaver!

$\int \frac{3}{\sqrt{4-3x^2}}\,dx$

I know I'm supposed to use the substitution $x= a\sin\theta$
Look ahead a step or two . . .

We want that radical to change from $\sqrt{4-3x^2}$

. . to: $\sqrt{4-4\sin^2\!\theta} \;=\;\sqrt{4(1-\sin^2\!\theta)} \;=\;\sqrt{4\cos^2\!\theta} \;=\;2\cos\theta$

To do that, we let: . $3x^2 \:=\:4\sin^2\!\theta \quad\Rightarrow\quad x^2 \:=\:\tfrac{4}{3}\sin^2\theta \quad \Rightarrow\quad x \:=\:\tfrac{2}{\sqrt{3}}\sin\theta$ .[1]

. . Then: . $dx \:=\:\tfrac{2}{\sqrt{3}}\cos\theta\,d\theta$

Substitute: . $\int\frac{3}{2\cos\theta}\left(\tfrac{2}{\sqrt{3}} \cos\theta\,d\theta\right) \;=\;\sqrt{3}\int d\theta \;=\;\sqrt{3}\,\theta + C$ .[2]

From [1], we have: . $\sin\theta \:=\:\tfrac{\sqrt{3}}{2}\,x \quad\Rightarrow\quad \theta \:=\:\arcsin\left(\tfrac{\sqrt{3}}{2}x\right)$

. . Substitute into [2]: . $\sqrt{3}\,\arcsin\left(\tfrac{\sqrt{3}}{2}\,x\righ t) + C$