We can use trigonometric identities to substitute things that even don’t look like trig.
If you see 1+x^2, tangent is probably useful. If you see 1-x^2, is probably useful.
Example 1
\begin{align*} \int \frac{1}{x^2+1} dx\\ x = \tan\theta\\ dx= \sec^2\theta d\theta\\ \int \frac{1}{\tan^2\theta+1} \sec^2\theta d\theta\\ \int \frac{\sec^2\theta}{\sec^2\theta} d\theta\\ \int 1 d\theta\\ = \theta + C\\ \text{because , then }\\ = tan^{-1} x + C\\ \end{align*}
Example 2
\begin{align*} &\int \frac{1}{\sqrt{1-x^2}} dx\\ x &= \sin \theta\\ dx = \cos \theta d\theta\\ &\int \frac{1}{\sqrt{1-\sin^2(\theta)}} \cos\theta d\theta\\ &\int \frac{1}{\sqrt{cos^2(\theta)}} \cos\theta d\theta\\ \text{note below}\\ \int \frac{\cos\theta}{\cos\theta}d\theta\\ \int 1 d\theta\\ = \theta + C\\ = sin^{-1}(x) + C\\ \end{align*}
Right, so isn’t necessarily the same as . It’s actually the absolute value, b/c , so it’s not necessarily a direct translation. In our case, we know that is always positive because it’s the inverse sine of x, which is always a positive number.
Example 3
\begin{align*} \int \sqrt{1-x^2}dx\\ x=\sin\theta\\ dx = \cos\theta d\theta\\ \int \sqrt{1-\sin^2\theta} cos\theta d\theta\\ \int \sqrt{cos^2\theta} cos\theta d\theta\\ \int cos\theta cos\theta d\theta\\ \int cos^2\theta d\theta\\ \int \frac{1+\cos(2\theta}{2} + C\\ \frac{1}{2} + \frac{1}{2}\cos(2\theta) d\theta\\ \frac{1}{2}\theta + \frac{1}{4} \sin(2\theta) + C\\ \text{lost it here..}\\ \frac{1}{2}sin^{-1}(x) + \frac{1}{2}x \sqrt{1-x^2} + C \end{align*}