Stackers (I tried to be nifty),
I'm currently in a Calculus II course, and we're working through the very basic steps of integrating. I'm really struggling with multiple portions of the substitution method that incorporates the use of u and du. I'm looking at this problem: $$\int (x^6-3x^2)^4(x^5-x)dx$$ Now comes my dilemma. I just cannot, for the life of me, understand how the substitution method would work with this equation. I just need someone to break it down for me a bit better than the book is, I'm supposed to find a value of u that helps the equation look like: $$\int u^4du$$ or something, right? If someone could just step through this problem with me, it'll help me greatly with the other 50 questions I have to do with the same concept. Is there any sort of fool-proof analysis that I can do to each problem to know where and how to assign the values of u and du?
Thanks for your help.
You don't like the power of something complicated (in particular, you do not want to have to expand the power). So you want the inside of the power to be $u$. (This is a good thing to try in other problems; depending on the setup, it may or may not work.) Then $u'=6x^5-6x$. You pretty much have this already, except for the $6$, but you can bring in constant multiples as you wish, provided you balance them. Specifically, you can write $\int (x^6-3x^2) (x^5-x) dx = \frac{1}{6} \int (x^6-3x^2) (6x^5-6x) dx$. Then with $u=(x^6-3x^2)$, your integral is $\frac{1}{6} \int u^4 du$.