How to Integrate a Polynomial Under a Square Root?

[LinearAlgebra]

How do you integrate this??

Integral of 3x^2 + (4-x^2)^(1/2) dx ??

I tried a u substitution for the 4-x^2 but what do you do with 3x^2? If someone could walk we through this, i'd greatly appreciate it...i hate being stuck on something so trivial in the middle of a problem.

 

[theperthvan]

Break it into two integrals, 3x^2 and (4-x^2)^(1/2).

Then use maybe x=2sin(u)

 

[LinearAlgebra]

sin? where did sin come into the picture? Is that the only way to do this?

 

[z-component]

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

Yes, break into two integrals:

\int 3x^2 \, dx + \int (4-x^2)^\frac{1}{2} \, dx

I think theperthvan is saying the second integral needs trig substitution.

 

[LinearAlgebra]

so is the second part equal to (2/3)(4-x^2)^3/2 ?? you don't have to do anything with the internal part, 4-x^2 like you do with the product rule in differentiation?

 

[z-component]

No, you cannot integrate like the product rule. The product rule is only for differentiation. The second integral must use trig substitution, I believe.

http://en.wikipedia.org/wiki/Trig_substitution

 

[arunma]

so is the second part equal to (2/3)(4-x^2)^3/2 ?? you don't have to do anything with the internal part, 4-x^2 like you do with the product rule in differentiation?

No, that will most definitely not work. Here you have a polynomial under a square root. You need to get rid of the square root. Here's what you do.

For \int \sqrt{a^{2}-x^{2}} \,dx, use x = a \, sin(\theta)

Note that you also need to substitute the differential, dx = a \, cos(\theta)d\theta

I hope I'm not saying too much, but also remember to use a certain trigonometric identity to eliminate the radical.

Having said this, if you don't know what a trigonometric substitution is, then my guess is that your calculus class hasn't yet covered it. Do you need to compute an antiderivative, or are you trying to compute a definite integral? Because if you're doing the definite integral from 0 to 2, you can do this simply by using the formula for the area of a circle.