trigonometric substitution

calculus

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, $1-si{n}^2(x)=co{s}^2(x)$ 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 $x=\tan \theta$, then ${\tan }^{-1}(x)=\theta$}\ = 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 $\sqrt{co{s}^2(\theta )}$ isn’t necessarily the same as $\cos \theta$. It’s actually the absolute value, b/c $\sqrt{x^2}\ne x,\sqrt{x^2}=|x|$, so it’s not necessarily a direct translation. In our case, we know that $\theta$ 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*}