How do the limits change with the substitution $u=\sin θ$

Question

Something very basic seems to slip my mind. Let's say that one wants to calculate the integral

$$\int_0^{2\pi}(1 + \sin \theta) \ d \theta$$ which is obviously equal to $2 \pi$.

Suppose that they, for whatever reason, want to use the substitution $u=\sin \theta$. If one tries to naively change the limits of integration, then both the lower and upper limits become $0$, which is wrong, since then the integral would vanish, which is not the case.

Where have they gone wrong?

Answer 1

When substituting $\theta$ with $u = f(\theta)$, f must be a 1-1 function over the integration interval.

Answer 2

What happens is that the relation between $d\theta$ and $du$ is not identical throughout the range $[0, 2\pi]$.

When you let $u$ equal $\sin \theta$, you have $\frac{du}{d \theta} = \cos \theta$, which will equal $\sqrt{1 - u^2}$ or $-\sqrt{1 - u^2}$ in different subranges of $[0, 2\ \pi]$. So, assuming you wanted to perform that substitution, you would have to split your integration as follows:

$$\int_0^{2\pi} (1 + \sin \theta) \; d\theta = \int_0^{\pi \over 2} (1 + \sin \theta) \; d\theta + \int_{\pi \over 2}^{3 \pi \over 2} (1 + \sin \theta) \; d\theta + \int_{3 \pi \over 2}^{2 \pi} (1 + \sin \theta) \; d\theta \\ = \int_0^1 {{1 + u} \over \sqrt{1 - u^2}} \; du - \int_1^{-1} {{1 + u} \over \sqrt{1 - u^2}} \; du + \int_{-1}^0 {{1 + u} \over \sqrt{1 - u^2}} \; du $$