# Thread: Trignometric Integrals

1. ## Trignometric Integrals

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

$\displaystyle \int sin^2(x)cos^3(x) dx$

I cant figure these 2 out.
Can you also show the steps?
Thank You so much

2. I would use u substitution on the first one using:
$\displaystyle u = 4-x^2$

and for the second one, try integration by parts:
$\displaystyle \int udv = uv - \int vdu$

3. Originally Posted by larryboi7
$\displaystyle \int \frac{x^2}{\sqrt{4-x^2}} dx$

$\displaystyle \int sin^2(x)cos^3(x) dx$

I cant figure these 2 out.
Can you also show the steps?
Thank You so much
use substitution method

HINT :
1) let x= 2 sin t
use sin^2(x)=1-cos^2(x)
2) $\displaystyle \int sin^2(x)cos^3(x) dx =\int sin^2(x) cos^2(x) cos(x)dx$
use cos^2(x)=1-sin^2(x) and let sin x=t so that cosx dx =dt and simplify

4. 1)

$\displaystyle x=2 \sin t \rightarrow dx=2 \cos t dt$

5. how do i find sin2t?

6. For the 2nd one, use cos(x)^2=1 - sin(x)^2
Then let u = sin x, and the remaining cos x is the du.

7. Originally Posted by danielomalmsteen
1)

$\displaystyle x=2 \sin t \rightarrow dx=2 \cos t dt$
after doing all that, how do i solve for sin2t?

8. Originally Posted by larryboi7
after doing all that, how do i solve for sin2t?
to restore the changes of variables?

9. Originally Posted by danielomalmsteen
to restore the changes of variables?
yes

10. $\displaystyle \sin {2t} = 2 \cdot \sin t \cdot \cos t = 2x \cdot \sqrt{1- \sin^2 t} = 2x \cdot \sqrt{1-\frac{x^2}{4}} = x \cdot \sqrt{4-x^2}$