# Do we require an invertible function in substitution method for integration?

When we use a substitution $$u = g(x)$$ to compute an indefinite integral $$\int f(x) dx$$ or a definite integral $$\int_a^b f(x)dx$$, what is required of the function $$g$$ in each case? Here are what I think. Please correct me if I am wrong.

For some integrals, we need to express $$x$$ in terms of $$u$$; in that case, I think we should require $$g$$ to be invertible on $$\mathbb{R}$$ (in the case of $$\int f(x) dx$$), or on $$[a, b]$$ (in the case of $$\int_a^b f(x)dx$$). Or else we need to discuss by cases, i.e. break the function $$g$$ into invertible pieces.

For integrals that do not require us to express $$x$$ in terms of $$u$$, consider for example $$\int \sin^2 x \cos x dx$$ and $$\int_0^{\frac{2\pi}{3}} \sin^2 x \cos x dx$$. Here $$u = \sin x$$ would work, but notice that the function $$g(x) = \sin x$$ is not invertible on $$[0, \frac{2\pi}{3}]$$, let alone on $$\mathbb{R}$$. So it seems to me that invertibility is not necessary.

In short, my question is about precisely which functions $$g$$ yield legitimate substitutions $$u = g(x)$$.

• When you do a $u$-substitution like $u = g(x)$, you don't care about invertibility. You just need the function to be differentiable on the interval of your given integral. Changing variables during integration doesn't inherently require monotonicity, invertibility, or injectivity except when obtaining the new integration limits after making an implicit substitution. You can also solve for $x$ as a function of $u$, but if you get that $x(u)$ is multivalued with more than one branch, you would have to sub differently along each branch. Commented Mar 6, 2023 at 1:49

$$g(x)$$ only has to be such that it is differenetiable.
I suppose there is also assumption that $$g'(x)$$ is an integrable function.