When integrating by subsitution, once you find the function $u$ and $du$, what function should I integrate?

Question

I understand that whenever we have a function to integrate, and we decide to use the method of variable change, what function should we integrate. For example, suppose this problem:

$ \int_{a}^{b}\frac{1}{\sqrt{1-x^{3}}}$ I decided that I will use the new function $u=\sqrt{1-x^{3}}$, and so $du=\frac{-3x^{2}}{2\sqrt{1-x^{3}}}$ or simply $\frac{-3x^{2}}{2u}$

Now, what goal should I keep in my mind when deciding what to put instead of the question mark here: $\int_{a}^{b}? du$ ? Should I just put $1/u$? Or something else?

Answer 1

First, let's change the integral into something that's actually solvable. $$\int\frac{x^2dx}{\sqrt{1-x^3}}$$ The goal of substitution is to get rid of all instances of $x$, including the $dx$ (which is missing in your question). For this integral, the $x^3$ and $x^2dx$ look promising, so let's use $$u=1-x^3$$ $$du = -3x^3dx \implies -\frac{1}{3}du = x^2dx$$ Substituting in $u$ and $du$: $$\int\frac{-\frac{1}{3}du}{\sqrt{u}} = -\frac{1}{3}\int u^{-1/2}du = -\frac{2}{3}u^{1/2} = -\frac{2}{3}\sqrt{u}$$ Reversing the substiution: $$ = -\frac{2}{3}\sqrt{1-x^3}$$ In order to integrate, all of the non-constant variables have to be the same. In this substitution, if $dx$ is not replaced and I get $$\int\frac{x^2dx}{\sqrt{u}},$$ then I can't do the integral since there are multiple moving parts: $x$ and $u$.

Answer 2

If you set $u=\sqrt{1-x^{3}}$, then $$du=\frac{-3x^{2}}{2\sqrt{1-x^{3}}} \color{red}{dx}.\tag1$$

Don't forget the $dx$ on the right-hand side! Also use $dx$ when writing the integral: $$\int_{a}^{b}\frac{1}{\sqrt{1-x^{3}}}\color{red}{dx}. $$ Consistent use of $dx$ will make the formal manipulations a lot clearer.

From Equation (1) we get $$dx=\frac{2\sqrt{1-x^{3}}}{-3x^{2}} du,$$ and we can use this equation to simply substitute $\dfrac{2\sqrt{1-x^{3}}}{-3x^{2}} du$ for $dx$ in the integral: $$ \int_{a}^{b}\frac{1}{\sqrt{1-x^{3}}} dx = \int_{a}^{b}\left(\frac{1}{\sqrt{1-x^{3}}}\right) \frac{2\sqrt{1-x^{3}}}{-3x^{2}} du = -\frac23\int_{a}^{b} \frac{1}{x^{2}} du. $$

That's correct, but not simple to solve, since we still need to convert the $x^2$ in the denominator into some function of $u$. It turns out $x^2 = (1 - u^2)^{2/3}$, so we end up with $$ -\frac23\int_{a}^{b} \frac{1}{(1 - u^2)^{2/3}} du, $$ which is not really an improvement over the original form of the integral. The obvious substitutions don't always work, and this is one of the cases where they do not.