Quick integration by substitiution question

[trap101]

integrate from (-1 to 1): [3(1-x²)²]/4 dx


so I'm having a little issue with the substitution for this. I'm going to let u = 1-x², which
can give me x² = 1-u, which I was going to put into my original equation and integrate. But I'm having issues finding the du to match with it.

 

[STEMucator]

integrate from (-1 to 1): [3(1-x²)²]/4 dxso I'm having a little issue with the substitution for this. I'm going to let u = 1-x², which
can give me x² = 1-u, which I was going to put into my original equation and integrate. But I'm having issues finding the du to match with it.

In this case you should actually expand to make the integrand easier and factor out the (3/4) :

$(1-x^2)^2 = x^4 - 2x^2 + 1$

 

[trap101]

In this case you should actually expand to make the integrand easier and factor out the (3/4) :

$(1-x^2)^2 = x^4 - 2x^2 + 1$

dang, your right. But now say I forgot that on a test. Is there anything I could do in other form to alleviate the situation, because I'm having a nightmare trying to figure something out.

 

[STEMucator]

dang, your right. But now say I forgot that on a test. Is there anything I could do in other form to alleviate the situation, because I'm having a nightmare trying to figure something out.

Take some deep breaths and ignore the problem for a few seconds, it will calm you down so you can think more clearly.

If you notice any obvious simplification you can do to the integrand before you actually evaluate it, then do it, sometimes a substitution isn't even required.

 

[Theorem.]

Zondrina is giving you the right idea here. Usually when you get stuck on these kind of problems you just have to stop and think: Where can I go with this? in your attempt you find that du=-2xdx, but there is no multiple of x in your integrand. This is problematic as you need to integrate with respect to u so you can't really take things further. What are your other options. One that almost instantly should come to mind with lower powers of polynomial terms like this one is just to expand. Once you expand a polynomial you can integrate term wise.