# Thread: Weird Integral

1. ## Weird Integral

A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

Upper Boundary = 2, Lower Boundary = 0

Integrate: (4-x)^(1/2).dx

Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?

2. You just need to solve $\displaystyle \int_{0}^{2}{\sqrt{4-x}\,dx}.$

Make the substitution $\displaystyle z^2=4-x,$ and don't forget to get the new integration limits.

3. Originally Posted by Evales
A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

Upper Boundary = 2, Lower Boundary = 0

Integrate: (4-x)^(1/2).dx

Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?
the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice

4. Originally Posted by Jhevon
the x is not squared. what you wrote does not require a trig sub. a substitution of u = 4 - x will suffice
Omg thanks for that. God your sharp yes the x is actually squared.

5. Originally Posted by Evales
A question that I have been given asks me to use a certain substitution to find the answer exactly. The only thing is not only do I think I can manage without it the substitution is the most random thing one I've seen.

Upper Boundary = 2, Lower Boundary = 0

Integrate: (4-x)^(1/2).dx

Use the substitution x = 2sin u to exactly evaluate this definite integral.
Should I just do it without using that substitution?
Originally Posted by Evales
Omg thanks for that. God your sharp yes the x is actually squared.
$\displaystyle \int_0^2 \sqrt{4-x^2}\,dx$

They tell you to use the substitution $\displaystyle x=2\sin u$. This implies that $\displaystyle dx=2\cos u \ \,du$

Change the limits of integration:

$\displaystyle 2=2\sin u \implies u=\arcsin(1)=\frac{\pi}{2}$
$\displaystyle 0=2\sin u \implies u=\arcsin(0)=0$

Thus, we have:

$\displaystyle \int_{0}^{\frac{\pi}{2}}2\cos u \sqrt{4-4\sin^2u}\,du$

Simplify the integrand:

$\displaystyle 2\cos u\sqrt{4-4\sin^2u}=4\cos u \sqrt{1-\sin^2u}=4\cos u\sqrt{\cos^2u}=4\cos^2u$

We now have:

$\displaystyle \int_{0}^{\frac{\pi}{2}}4\cos^2u\,du$

I'll let you take it from here, noting that $\displaystyle \cos^2u=\frac{1+\cos (2u)}{2}$.

Hope that this helped!