Chapter 46: Trigonometric Integrals

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


Example - sin5(x)cos(x)dx\int sin^{5}(x)cos(x)dx

Apply Substitution

u=sin(x)du=cos(x)dxu=sin(x) \Rightarrow du=cos(x)dx

sin5(x)cos(x)dx=u5du\int sin^{5}(x)cos(x)dx=\int u^{5}du

=16u6+C=\frac{1}{6}u^{6}+C

=16sin(6)(x)+C=\frac{1}{6}sin(6)(x)+C


Example - sin3(x)cos3(x)dx\int sin^{3}(x)cos^{3}(x)dx

=sin3(x)cos2(x)cos(x)dx=\int sin^{3}(x)\cdot cos^{2}(x)\cdot cos(x)dx

=sin3(x)[1sin2(x)]cos(x)dx=\int sin^{3}(x)[1-sin^{2}(x)]cos(x)dx

Apply Substitution

u=sin(x)du=cos(x)u=sin(x) \Rightarrow du=cos(x)

=u3(1u2)du=u3u5du=\int u^{3}(1-u^{2})du = \int u^{3}-u^{5}du

=14u416u6+C=14sin4(x)16sin6(x)+C=\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 cos2(x)+sin2(x)=12cos^{2}(x)+sin^{2}(x)=1^{2}


Example - sin5(x)cos4(x)dx\int sin^{5}(x)cos^{4}(x)dx

=sin(x)[sin4(x)]cos4(x)dx=\int sin(x)[sin^{4}(x)]cos^{4}(x)dx

=sin(x)[1cos2(x)]2cos4(x)dx=\int sin(x)[1-cos^{2}(x)]^{2}cos^{4}(x)dx

Apply substitution

u=cos(x)du=sin(x)dxu=cos(x) \Rightarrow du=-sin(x)dx

=[1cos2(x)]2cos4(x)(sinx)dx=-\int [1-\underbrace{cos^{2}}(x)]^{2}\underbrace{cos^{4}}(x)\underbrace{(-sinx)dx}

uu uu dudu

=[1u2]2u4du=(12u2+u4)u4du=-\int [1-u^{2}]^{2}u^{4}du=-\int (1-2u^{2}+u^{4})u^{4}du

=u42u6+u8du=15u527u7+19u9+C=-\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 dudu


Key Fact

sin2(x)=1cos(2x)2sin^{2}(x)=\frac{1-cos(2x)}{2}

cos2(x)=1+cos(2x)2cos^{2}(x)=\frac{1+cos(2x)}{2}


Example - sin2(x)cos2(x)dx\int sin^{2}(x)cos^{2}(x)dx

=(1cos2x2)(1+cos(2x)2)dx=\int (\frac{1-cos2x}{2})(\frac{1+cos(2x)}{2})dx

=1cos2(2x)4dx=1414cos2(2x)dx=\int \frac{1-cos^{2}(2x)}{4}dx= \int \frac{1}{4}-\frac{1}{4}cos^{2}(2x)dx

=1418(1+cos(4x))dx=\int \frac{1}{4}-\frac{1}{8}(1+cos(4x))dx

=14x18x132sin(4x)+C=\frac{1}{4}x-\frac{1}{8}x-\frac{1}{32}sin(4x)+C


Example - Calculate 0π41+cos(2x)dx\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx

1+cos(2x)2=cos2(x)1+cos(2x)=2cos2(x)\frac{1+cos(2x)}{2}=cos^{2}(x)\Rightarrow 1+cos(2x)= 2cos^{2}(x)

Thus, 0π41+cos(2x)dx\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx

The absolute value appears since x2=x\sqrt{x^{2}}=|x|

=0π42cos2(x)dx=\int_{0}^{\frac{\pi}{4}}\sqrt{2 cos^{2}(x)}dx

=0π42cos(2x)dx=\int_{0}^{\frac{\pi}{4}}\sqrt{2}|cos(2x)|dx

=0π42cos(2x)dx=\int_{0}^{\frac{\pi}{4}}\sqrt{2}cos(2x)dx

=[22sin(2x)]0π4=[\frac{\sqrt{2}}{2}sin(2x)]_{0}^{\frac{\pi}{4}}

=22sin(π2)22sin(0)=22=\frac{\sqrt{2}}{2}sin(\frac{\pi}{2})-\frac{\sqrt{2}}{2}sin(0)=\frac{\sqrt{2}}{2}


Discuss - 0π161cos(4x)dx\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx

1cos(4x)2=sin2(2x)1cos(4x)=2sin2(2x)\frac{1-cos(4x)}{2}=sin^{2}(2x) \Rightarrow 1- cos(4x)=2sin^{2}(2x)

Thus, 0π161cos(4x)dx=0π162sin2(2x)dx\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx=\int_{0}^{\frac{\pi}{16}}\sqrt{2sin^{2}(2x)}dx

=0π162sin(2x)dx=\int_{0}^{\frac{\pi}{16}}\sqrt{2}sin(2x)dx

=[22cos(2x)]0π16=[-\frac{\sqrt{2}}{2}cos(2x)]_{0}^{\frac{\pi}{16}}


Key Facts

sin2(x)+cos2(x)=1tan2(x)+1=sec2(x)sin^{2}(x)+cos^{2}(x)=1\Leftrightarrow tan^{2}(x)+1=sec^{2}(x)

Divide both sides by cos2(x)cos^{2}(x)

sec(x)tan(x)dx=sec(x)\int sec(x)tan(x)dx=sec(x) sec2(x)dx=tan(x)\int sec^{2}(x)dx=tan(x)


Example - tan4(x)dx\int tan^{4}(x)dx

=tan2(x)[sec2(x)1]dx=\int tan^{2}(x)[sec^{2}(x)-1]dx

=tan2(x)sec2(x)tan2(x)dx=\int tan^{2}(x)sec^{2}(x)-tan^{2}(x)dx

=tan2(x)sec2(x)sec2(x)+1dx=\int \underbrace{ tan^{2}}(x)\underbrace{sec^{2}}(x)-\underbrace{sec^{2}(x)+1dx}

uu dudu we can integrate these

=u2dusec2(x)dx+1dx=\int u^{2}du - \int sec^{2}(x)dx + \int 1dx

=13(tan(x))3tan(x)+x+C=\frac{1}{3}(tan(x))^{3}-tan(x)+x+C


Discuss - tan4(x)sec4(x)dx\int tan^{4}(x)sec^{4}(x)dx

=tan4(x)(1+tan2(x))sec2(x)dx=\int tan^{4}(x)\cdot (1+tan^{2}(x))sec^{2}(x)dx

=tan4(x)sec2(x)dx+tan6(x)sec2(x)dx=\int tan^{4}(x)sec^{2}(x)dx + \int tan^{6}(x)sec^{2}(x)dx

=15tan5(x)+17(x)+C=\frac{1}{5}tan^{5}(x)+\frac{1}{7}(x)+C


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