# Thread: Find the integral of a function

1. ## Find the integral of a function

I'm working the practice problems in my book, and I keep getting this question wrong. Can you show me the steps of how to solve this, please, because I have no idea of what to do? I changed the numbers, so I could try to solve the question in the book myself. I just want to see the steps.

∫(1-x²)⁴x dx

2. ## Re: Find the integral of a function

Try rewriting this as $\displaystyle \displaystyle -\frac{1}{2}\int{-2x(1 - x^2)^4\,dx}$. Does the required substitution become a bit more obvious this way?

3. ## Re: Find the integral of a function

Hello, RedonYellow!

$\displaystyle \int (1-x^2)^4\,x\,dx$

$\displaystyle \text{Let: }u \,=\,1-x^2 \quad\Rightarrow\quad du \,=\,\text{-}2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\text{-}\tfrac{1}{2}du$

$\displaystyle \text{Substitute: }\;\int u^4\left(\text{-}\tfrac{1}{2}\,du\right) \;=\;\text{-}\tfrac{1}{2}\int u^4\,du$

Got it?

4. ## Re: Find the integral of a function

Of course, you could have noted that $\displaystyle (a+ b)^4= a^4+ 4a^3b+ 6a^2b^2+ 4a^3b+ b^4$ so that $\displaystyle (1- x^2)^4= 1- 4x^2+ 6x^4- 4x^6+ x^8$ and then $\displaystyle x(1- x^2)^4= x- 4x^3+ 6x^5- 4x^7+ x^9$. But, yes, it is easier to use the substitution.