# Thread: trig sub integral?

1. ## trig sub integral?

so, the problem is integral sq rt (1-4x^2)dx

i really didn't know how to start, other than the fact that it looks sort of like the a^2-x^2 form of trig sub (in which i think you're supposed to make x= asintheta)

so, i tried that, but i got lost/confused not long after

2. Hello, buttonbear!

You have the right idea . . . just carry it through.

$\int \sqrt{1-4x^2}\,dx$

Let: . $2x = \sin\theta \quad\Rightarrow\quad x = \tfrac{1}{2}\sin\theta \quad\Rightarrow\quad dx = \tfrac{1}{2}\cos\theta\,d\theta$

Substitute: . $\int(\cos\theta)\left(\tfrac{1}{2}\cos\theta\,d\th eta\right) \;=\;\tfrac{1}{2}\int\cos^2\!\theta\,d\theta \;=\;\tfrac{1}{2}\int\frac{1+\cos2\theta}{2}\,d\th eta \;=\;\tfrac{1}{4}\int(1 + \cos2\theta)\,d\theta$

And we have: . $\tfrac{1}{4}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{4}\left(\theta + \tfrac{1}{2}[2\sin\theta\cos\theta]\right) + C$

. . . . . . . . . $= \;\tfrac{1}{4}(\theta + \sin\theta\cos\theta) + C$

Back-substitute: . $\sin\theta \:=\:\frac{2x}{1} \:=\:\frac{opp}{hyp}$

. . Then: . $adj \,=\,\sqrt{1-4x^2} \quad\Rightarrow\quad \cos\theta \:=\:\sqrt{1-4x^2}$

Theefore: . $\frac{1}{4}\left(\arcsin2x + 2x\sqrt{1-4x^2}\right) + C$

3. thanks so much for your help..would you mind just explaining where that first cos comes from? in the integral?

4. Originally Posted by buttonbear
thanks so much for your help..would you mind just explaining where that first cos comes from? in the integral?
When substituting $x=a\sin\theta,$ the radical $\sqrt{a^2-x^2}$ becomes $a\cos\theta.$ Here is why:

$\sqrt{a^2-x^2}$

$=\sqrt{a^2-(a\sin\theta)^2}$

$=\sqrt{a^2-a^2\sin^2\theta}$

$=\sqrt{a^2(1-\sin^2\theta)}$

$=\sqrt{a^2\cos^2\theta}$

$=a\cos\theta$ (assuming $a\geq0$ and a suitable restriction on $\theta$)