Indefinite Integrals
This is messy. However, notice that is the derivative of . So if we substitute for , then we take this function from the word into the world.
u &= x^{2} \\ du &= 2x\ dx \\ \end{align}$$ Substitute these into the original equation. $$\int \cos(u) \, du $$ Now that is a lot simpler to solve. Substitute back the $x$ for $u$ after you are done. $$\int \cos(x^{2})2x \, dx $$ # Definite Integrals If our old function was a definite integral, we would need more work. $$\int_{1}^2 2x\cos (x^2) \, dx $$ There are two ways to go about this. ### Case 1: Substitute bounds After $u$-substitution, we get $$\int \cos(u) \, du $$ We cannot put the bounds 1 and 2 though, since they are in respect to $x$. Instead, see what $u$ is when $x$ is at that value. $u = x^2 = 1^2 = 1$. Put these into the bounds, and evaluate that way. ### Case 2: Substitute indefinite integral The other approach is to continue with normal indefinite integrals, obtain the expression in terms of $x$, and use the same bounds. For this function, it would be $$\int_{1}^2 \cos(x^{2})2x \, dx $$ ## Special cases There are times u-substitution is not necessary but can simplify the expression. For example, $$(x-1)(x-3)^4$$ can turn into $$(u+2)u^{4}$$ [Khan Article](https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-9/a/review-applying-u-substitution)