# Can we use trig substitution with cosine?

From what I've gathered, every text book says we can do trig substitution with either sin, sec or tan. Some other examples use tan(x/2), but so far I've never encountered cosine substitution. Is there some reason why it's omitted, or is it just interchangeable with sin? https://en.wikipedia.org/wiki/Trigonometric_substitution

$$x=\sin \theta: \sqrt{1-x^2}=\sqrt{1-\sin^2 \theta}=\sqrt{\cos^2 \theta}=\cos(\theta)$$

$$x=\cos \theta:\sqrt{1-x^2}=\sqrt{1-\cos^2 \theta}=\sqrt{\sin^2 \theta}=\sin(\theta)$$

From the example above, it looks like I could use either when integrating.

• It is interchangeable with sine. Oct 24, 2021 at 1:25
• As long as it is a strictly increasing/decreasing continuous function you're using for the substitution, then yes it works Oct 24, 2021 at 1:33
• Write $x = \sin \theta$ with \theta instead, and $\cos \theta, \sin \theta$ as \cos \theta, \sin \theta. For more, you can see the MathJax reference. Oct 24, 2021 at 2:37

Like you said, both $$\cos \theta$$ and $$\sin \theta$$ work. The domains are a little different: $$\sin \theta$$ is one-to-one when $$t \in [-\pi/2, \pi/2]$$ and so on, whereas $$\cos \theta$$ is one-to-one when $$t \in [-\pi, 0]$$ or $$[0, \pi]$$.

However, $$x = \cos \theta$$, $$dx = -\sin \theta \ d \theta$$ so you have an extra negative sign. If you forget the negative sign, you might spend ages poring over your work.