>[!warning] trigonometry jumpscare Trigonometric Substitution is a way to solve an [[Indefinite Integral]] by leveraging [[Trigonometric Identities]] when substituting $x$. For when $a = \text{const}.$ $\huge\begin{align*} \int\sqrt{a^{2}-x^{2}\mathrm{d}x}\\ x=a\sin\theta \end{align*}$ $\huge\begin{align*} \int\sqrt{a^{2}+x^2}\mathrm{d}x \\ x=a\tan\theta \end{align*}$ $\huge\begin{align*} \int\sqrt{x^2-a^{2}\mathrm{d}x}\\ x =a\sec\theta \end{align*}$ >[!example] >$ >\int \frac{\sqrt{4-x^2}}{x^{2}}\mathrm{d}x >$ >$\begin{align*} a^{2}&= 4 \\ a &= \sqrt{4}= 2 \\ x &= a\sin\theta = 2\sin\theta\\ \mathrm{d}x &= 2\cos\theta \mathrm{d}\theta \\ \int \frac{\sqrt{4-x^2}}{x^{2}}\mathrm{d}x &= \int \frac{ \sqrt{4 - (2\sin\theta)^2} }{(2\sin\theta)^2} 2\cos(\theta)\mathrm{d}\theta \\ &= \int \frac{ \sqrt{4 - 4\sin^2\theta} }{4\sin^2\theta} 2\cos(\theta)\mathrm{d}\theta \\ &= \int \frac{ 2\sqrt{1 - \sin^2\theta} }{2\sin^2\theta} \cos(\theta)\mathrm{d}\theta \\ &= \int \frac{ 2\cos\theta }{2\sin^2\theta} \cos(\theta)\mathrm{d}\theta \\ &= \int \frac{ \cos^2\theta }{\sin^2\theta} \mathrm{d}\theta \\ &= \int \frac{ 1-\sin^2\theta }{\sin^2\theta} \mathrm{d}\theta \\ &= \int \frac{ 1-\sin^2\theta }{\sin^2\theta} \mathrm{d}\theta \\ &= \int \left( \frac{1}{\sin^2\theta}- \frac{\sin^2\theta}{\sin^2\theta} \right)\mathrm{d}\theta \\ &= \int \left( \csc^{2}\theta - 1 \right) \mathrm{d}\theta \\ &= -\cot \theta - \theta + C \\ x &= 2\sin\theta \\ \frac{x}{2}&= \sin \theta \\ \tan \theta &= \frac{x}{\sqrt{4-x^2}}\\ \cot\theta &= \frac{\sqrt{4-x^2}}{x} \\ \sin^{-1} \frac{x}{2} &= \theta \\ -\cot \theta - \theta + C &= -\frac{\sqrt{4-x^{2}}}{x} - \sin^{-1} \frac{x}{2}+C \\ \end{align*}$ >$ \int \frac{\sqrt{4-x^2}}{x^{2}}\mathrm{d}x = -\frac{\sqrt{4-x^{2}}}{x} - \sin^{-1} \frac{x}{2}+C $