## Sin & Cos The typical strategy for [[Integration|integrating]] [[Odd]] powers of [[Trigonometry|Trigonometric Functions]] is to [[Factor]] a sin or cosine out, then use the [[Pythagorean Identity]] and substitution. >[!example] >$\huge \int \sin ^3 (x) \d x$ >>[!check]- Solution >>$\huge \begin{align} >>\int \sin ^3 (x) \d x &= >>\int \sin(x)\sin ^2 (x) \d x \\ >>&= >>\int \sin(x)\pa{1-\cos^2 x} \d x \\ >>\\ >>\let u &=\cos(x) \\ >>\let \d u &= -\sin(x)\d x \\ >>\\ >>\int \sin(x)\pa{1-\cos^2 x} \d x &= >>\int -\pa{1-u^2} \d u \\ >>&= \int\pa{u^2-1}\d u \\ >>&= \frac{1}{3} u^{3} - u + C\\ >>&= \boxed{\frac{1}{3} \cos^3 x - \cos x + C} >>\end{align}$ ## Tangeant >[!example] >$\huge \int \tan^3 (x) \d x$ >>[!check]- Solution >>$\huge \begin{align} >>\int \tan^3 x \d x &= >>\int \tan(x)\tan^2(x) \d x \\ >>&= >>\int \tan(x)\pa{\sec^2(x) - 1} \d x \\ &= >>\int\tan(x)\sec^2(x)\d x - \int\tan (x)\d x \\ &= >>\int\tan(x)\sec^2(x)\d x - \ln \left| \sec x\right| + C \\ >>\\ >>\let u &=\tan(x) \\ >>\d u &= \sec^2(x)\d x\\ >> >>\cdots &= >>\int u \d u - \ln \left| \sec x\right| + C \\ >>&= \frac{1}{2}u^2 - \ln \left| \sec x\right| + C \\ >>&= \frac{1}{2}\tan^2(x) - \ln \left| \sec x\right| + C \\ >> >>\end{align}$ >[!example] >$\huge \int \sec (x)\d x $ >>[!check]0 Solution >>$\huge \begin{align} >>\int \sec (x)\d x &= >>\int \pa{ \sec (x) >>\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} >>}\d x >>\\&=\int { >>\frac{\sec^2(x)+\sec (x)\tan(x)}{\sec(x)+\tan(x)} >>}\d x >>\\ >>\let u &= \sec(x)+\tan(x) \\ >>\let \d u &= (\sec(x)\tan(x) + \sec^2(x))\d x >>\\ >>\\ >>\cdots &= \int \frac{1}{u} \d u \\ >>&= \ln \left| u \right| + C\\ >>&= \boxed{ >> \ln \left| \sec(x)+\tan(x) \right| + C >>} >>\end{align}$