# Chapter 46: Trigonometric Integrals

We introduce a variety of techniques for dealing with trigonometric functions in integrals

Example - ﻿$\int sin^{5}(x)cos(x)dx$﻿

Apply Substitution

﻿$u=sin(x) \Rightarrow du=cos(x)dx$﻿

﻿$\int sin^{5}(x)cos(x)dx=\int u^{5}du$﻿

﻿$=\frac{1}{6}u^{6}+C$﻿

﻿$=\frac{1}{6}sin(6)(x)+C$﻿

Example - ﻿$\int sin^{3}(x)cos^{3}(x)dx$﻿

﻿$=\int sin^{3}(x)\cdot cos^{2}(x)\cdot cos(x)dx$﻿

﻿$=\int sin^{3}(x)[1-sin^{2}(x)]cos(x)dx$﻿

Apply Substitution

﻿$u=sin(x) \Rightarrow du=cos(x)$﻿

﻿$=\int u^{3}(1-u^{2})du = \int u^{3}-u^{5}du$﻿

﻿$=\frac{1}{4}u^{4}-\frac{1}{6}u^{6}+C=\frac{1}{4}sin^{4}(x)-\frac{1}{6}sin^{6}(x)+C$﻿

Use one trig function for the substitution. Convert extra cos terms to sin terms using ﻿$cos^{2}(x)+sin^{2}(x)=1^{2}$﻿

Example - ﻿$\int sin^{5}(x)cos^{4}(x)dx$﻿

﻿$=\int sin(x)[sin^{4}(x)]cos^{4}(x)dx$﻿

﻿$=\int sin(x)[1-cos^{2}(x)]^{2}cos^{4}(x)dx$﻿

Apply substitution

﻿$u=cos(x) \Rightarrow du=-sin(x)dx$﻿

﻿$=-\int [1-\underbrace{cos^{2}}(x)]^{2}\underbrace{cos^{4}}(x)\underbrace{(-sinx)dx}$﻿

﻿$u$﻿ ﻿$u$﻿ ﻿$du$﻿

﻿$=-\int [1-u^{2}]^{2}u^{4}du=-\int (1-2u^{2}+u^{4})u^{4}du$﻿

﻿$=-\int u^{4}-2u^{6}+u^{8}du=-\frac{1}{5}u^{5}-\frac{2}{7}u^{7}+\frac{1}{9}u^{9}+C$﻿

Use the odd power to produce ﻿$du$﻿

#### Key Fact

﻿$sin^{2}(x)=\frac{1-cos(2x)}{2}$﻿

﻿$cos^{2}(x)=\frac{1+cos(2x)}{2}$﻿

Example - ﻿$\int sin^{2}(x)cos^{2}(x)dx$﻿

﻿$=\int (\frac{1-cos2x}{2})(\frac{1+cos(2x)}{2})dx$﻿

﻿$=\int \frac{1-cos^{2}(2x)}{4}dx= \int \frac{1}{4}-\frac{1}{4}cos^{2}(2x)dx$﻿

﻿$=\int \frac{1}{4}-\frac{1}{8}(1+cos(4x))dx$﻿

﻿$=\frac{1}{4}x-\frac{1}{8}x-\frac{1}{32}sin(4x)+C$﻿

Example - Calculate ﻿$\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx$﻿

﻿$\frac{1+cos(2x)}{2}=cos^{2}(x)\Rightarrow 1+cos(2x)= 2cos^{2}(x)$﻿

Thus, ﻿$\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx$﻿

The absolute value appears since ﻿$\sqrt{x^{2}}=|x|$﻿

﻿$=\int_{0}^{\frac{\pi}{4}}\sqrt{2 cos^{2}(x)}dx$﻿

﻿$=\int_{0}^{\frac{\pi}{4}}\sqrt{2}|cos(2x)|dx$﻿

﻿$=\int_{0}^{\frac{\pi}{4}}\sqrt{2}cos(2x)dx$﻿

﻿$=[\frac{\sqrt{2}}{2}sin(2x)]_{0}^{\frac{\pi}{4}}$﻿

﻿$=\frac{\sqrt{2}}{2}sin(\frac{\pi}{2})-\frac{\sqrt{2}}{2}sin(0)=\frac{\sqrt{2}}{2}$﻿

Discuss - ﻿$\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx$﻿

﻿$\frac{1-cos(4x)}{2}=sin^{2}(2x) \Rightarrow 1- cos(4x)=2sin^{2}(2x)$﻿

Thus, ﻿$\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx=\int_{0}^{\frac{\pi}{16}}\sqrt{2sin^{2}(2x)}dx$﻿

﻿$=\int_{0}^{\frac{\pi}{16}}\sqrt{2}sin(2x)dx$﻿

﻿$=[-\frac{\sqrt{2}}{2}cos(2x)]_{0}^{\frac{\pi}{16}}$﻿

#### Key Facts

﻿$sin^{2}(x)+cos^{2}(x)=1\Leftrightarrow tan^{2}(x)+1=sec^{2}(x)$﻿

Divide both sides by ﻿$cos^{2}(x)$﻿

﻿$\int sec(x)tan(x)dx=sec(x)$﻿ ﻿$\int sec^{2}(x)dx=tan(x)$﻿

Example - ﻿$\int tan^{4}(x)dx$﻿

﻿$=\int tan^{2}(x)[sec^{2}(x)-1]dx$﻿

﻿$=\int tan^{2}(x)sec^{2}(x)-tan^{2}(x)dx$﻿

﻿$=\int \underbrace{ tan^{2}}(x)\underbrace{sec^{2}}(x)-\underbrace{sec^{2}(x)+1dx}$﻿

﻿$u$﻿ ﻿$du$﻿ we can integrate these

﻿$=\int u^{2}du - \int sec^{2}(x)dx + \int 1dx$﻿

﻿$=\frac{1}{3}(tan(x))^{3}-tan(x)+x+C$﻿

Discuss - ﻿$\int tan^{4}(x)sec^{4}(x)dx$﻿

﻿$=\int tan^{4}(x)\cdot (1+tan^{2}(x))sec^{2}(x)dx$﻿

﻿$=\int tan^{4}(x)sec^{2}(x)dx + \int tan^{6}(x)sec^{2}(x)dx$﻿

﻿$=\frac{1}{5}tan^{5}(x)+\frac{1}{7}(x)+C$﻿