# Thread: integral - trig substitution?

1. ## integral - trig substitution?

Is this a trig substitution or a u-substitution problem?
It dont seem to work out for trig substitution..

$\displaystyle \int \frac {x^3}{ \sqrt {1 - 4x^2}}dx$

$\displaystyle x = sin\theta$

$\displaystyle dx = cos\theta d\theta$

$\displaystyle \sqrt{1-4x^2}$

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

$\displaystyle = 2cos\theta$

sub back in

$\displaystyle \int \frac {sin^3\theta}{2 cos\theta}$

is this correct?

2. ## Re: integral - trig substitution?

$\displaystyle \int \frac {x^3}{ \sqrt {1 - 4x^2}} \, dx$

$\displaystyle u = 1-4x^2$

$\displaystyle du = -8x \, dx$

$\displaystyle x^2 = \frac{1-u}{4}$

$\displaystyle -\frac{1}{8} \int \frac {x^2 \cdot (-8x)}{ \sqrt {1 - 4x^2}} \, dx$

$\displaystyle -\frac{1}{32} \int \frac{1-u}{\sqrt{u}} \, du$

$\displaystyle -\frac{1}{32} \int \frac{1}{\sqrt{u}} - \sqrt{u} \, du$

3. ## Re: integral - trig substitution?

thank you i got it now

4. ## Re: integral - trig substitution?

You finish with the power rule. Square roots correspond to exponents of (1/2). Exponents in the denominator correspond to negative exponents in the numerator.

5. ## Re: integral - trig substitution?

Originally Posted by icelated
How do you finish that? because i have no idea =(
are you saying that you are unable to find the antiderivative of

$\displaystyle -\frac{1}{32} \int \frac{1}{\sqrt{u}} - \sqrt{u} \, \, du$

and back-substitute $\displaystyle (1 - 4x^2)$ for $\displaystyle u$ ?

6. ## Re: integral - trig substitution?

thank you i got it finished

7. ## Re: integral - trig substitution?

Originally Posted by icelated
thank you i got it finished
Good deal. I think skeeter and I were starting to get worried!