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Integration II

Integration with trigonometric identities

Integration with trigonometric identities

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Explainer Video

Tutor: Bilal

Summary

Integration with trigonometric identities

In a nutshell

Many integrals can be simplified by using any number of trigonometric identities.



List of trigonometric identities

Here is a list of all the trigonometric identities you need to know.


description

identity

Basic definitions

tan⁡(x)≡sin⁡(x)cos⁡(x)\tan(x)\equiv \dfrac{\sin(x)}{\cos(x)}tan(x)≡cos(x)sin(x)​​​

​cosec⁡(x)≡1sin⁡(x)\cosec(x)\equiv \dfrac{1}{\sin(x)}cosec(x)≡sin(x)1​​​

​sec⁡(x)≡1cos⁡(x)\sec(x) \equiv \dfrac{1}{\cos(x)}sec(x)≡cos(x)1​​​​

cot⁡(x)≡1tan⁡(x)≡cos⁡(x)sin⁡(x)\cot(x)\equiv \dfrac{1}{\tan(x)}\equiv \dfrac{\cos(x)}{\sin(x)}cot(x)≡tan(x)1​≡sin(x)cos(x)​​​

Basic identities

sin⁡2(x)+cos⁡2(x)≡1\sin^2(x)+\cos^2(x)\equiv 1sin2(x)+cos2(x)≡1​​

​sec⁡2(x)≡1+tan⁡2(x)\sec^2(x)\equiv 1 + \tan^2(x)sec2(x)≡1+tan2(x)​​

​cosec⁡2(x)≡1+cot⁡2(x)\cosec^2(x) \equiv 1+ \cot^2(x)cosec2(x)≡1+cot2(x)​​

Compound angle identities

​​​sin⁡(A+B)≡sin⁡(A)cos⁡(B)±cos⁡(A)sin⁡(B)\sin(A+B)\equiv \sin(A)\cos(B)\pm \cos(A)\sin(B)sin(A+B)≡sin(A)cos(B)±cos(A)sin(B)​​

​cos⁡(A±B)≡cos⁡(A)cos⁡(B)∓sin⁡(A)sin⁡(B)\cos(A\pm B)\equiv \cos(A)\cos(B)\mp\sin(A)\sin(B)cos(A±B)≡cos(A)cos(B)∓sin(A)sin(B)​​

tan⁡(A+B)=tan⁡(A)±tan⁡(B)1∓tan⁡(A)tan⁡(B)\tan(A+B)=\dfrac{\tan(A)\pm \tan(B)}{1\mp \tan(A)\tan(B)}tan(A+B)=1∓tan(A)tan(B)tan(A)±tan(B)​​​

​​Double angle identities

sin⁡(2x)≡2sin⁡(x)cos⁡(x)\sin(2x)\equiv 2\sin(x)\cos(x)sin(2x)≡2sin(x)cos(x)​​

​cos⁡(2x)≡cos⁡2(x)−sin⁡2(x)≡2cos⁡2(x)−1≡1−2sin⁡2(x)\begin{aligned}\cos(2x)&\equiv \cos^2(x)- \sin^2(x)\\&\equiv2\cos^2(x)-1\\ &\equiv 1-2\sin^2(x)\end{aligned}cos(2x)​≡cos2(x)−sin2(x)≡2cos2(x)−1≡1−2sin2(x)​​​

​tan⁡(2x)≡2tan⁡(x)1−tan⁡2(x)\tan(2x)\equiv \dfrac{2\tan(x)}{1-\tan^2(x)}tan(2x)≡1−tan2(x)2tan(x)​​​


These can be used to make certain integrals much simpler.


Example 1

Compute the following integrals:

i) ∫cos⁡2(x) dx\int \cos^2(x)\,dx∫cos2(x)dx​

ii) ∫1cosec⁡(2x)(cos⁡(x)sin⁡(x)+sin⁡(x)cos⁡(x)) dx\int \dfrac{1}{\cosec(2x)}\left(\dfrac{\cos(x)}{\sin(x)}+\dfrac{\sin(x)}{\cos(x)}\right)\, dx∫cosec(2x)1​(sin(x)cos(x)​+cos(x)sin(x)​)dx​


Part i):

Use the double angle formula to express cos⁡2(x)\cos^2(x)cos2(x) in terms of cos⁡(2x)\cos(2x)cos(2x):

​cos⁡(2x)≡2cos⁡2(x)−1⇒cos⁡2(x)≡cos⁡(2x)+12\begin{aligned} \cos(2x)&\equiv 2\cos^2(x)-1\\ \Rightarrow \cos^2(x)&\equiv \dfrac{\cos(2x)+1}{2} \end{aligned}cos(2x)⇒cos2(x)​≡2cos2(x)−1≡2cos(2x)+1​​​​


This is easier to integrate:

∫cos⁡2(x) dx≡∫(cos⁡(2x)+12) dx=12∫cos⁡(2x) dx+∫12 dx=12(12sin⁡(2x))+12x+C=14sin⁡(2x)+12x+C\begin{aligned} \int \cos^2(x) \, dx&\equiv \int \left(\dfrac{\cos(2x)+1}{2}\right) \, dx\\&= \dfrac12\int \cos(2x)\, dx + \int\dfrac12 \, dx\\&=\dfrac12 \left(\dfrac12 \sin(2x)\right) + \dfrac12 x+C\\&=\dfrac14 \sin(2x) + \dfrac12x+C \end{aligned}∫cos2(x)dx​≡∫(2cos(2x)+1​)dx=21​∫cos(2x)dx+∫21​dx=21​(21​sin(2x))+21​x+C=41​sin(2x)+21​x+C​​​


​∫cos⁡2(x)=14sin⁡(2x)+12x+C‾ \int \cos^2(x) =\underline{\dfrac14 \sin(2x) + \dfrac12x+C}∫cos2(x)=41​sin(2x)+21​x+C​​​


Part ii):

Simplify the integrand by writing everything as one fraction:

∫1cosec⁡(2x)(cos⁡(x)sin⁡(x)+sin⁡(x)cos⁡(x)) dx≡∫1cosec⁡(2x)(cos⁡2(x)+sin⁡2(x)sin⁡(x)cos⁡(x)) dx \int \dfrac{1}{\cosec(2x)}\left(\dfrac{\cos(x)}{\sin(x)}+\dfrac{\sin(x)}{\cos(x)}\right)\, dx \equiv \int\dfrac{1}{\cosec(2x)}\left( \dfrac{\cos^2(x)+\sin^2(x)}{\sin(x)\cos(x)}\right)\,dx∫cosec(2x)1​(sin(x)cos(x)​+cos(x)sin(x)​)dx≡∫cosec(2x)1​(sin(x)cos(x)cos2(x)+sin2(x)​)dx​​


Use the following identities to simplify this:

cos⁡2(x)+sin⁡2(x)≡1sin⁡(2x)≡2sin⁡(x)cos⁡(x)⇒sin⁡(x)cos⁡(x)≡12sin⁡(2x)cosec⁡(x)≡1sin⁡(x)⇒cosec⁡(2x)≡1sin⁡(2x)\cos^2(x)+\sin^2(x) \equiv 1\\ \sin(2x)\equiv 2\sin(x)\cos(x) \Rightarrow \sin(x)\cos(x) \equiv \dfrac12 \sin(2x)\\ \cosec(x)\equiv \dfrac{1}{\sin(x)} \Rightarrow \cosec(2x) \equiv \dfrac{1}{\sin(2x)}cos2(x)+sin2(x)≡1sin(2x)≡2sin(x)cos(x)⇒sin(x)cos(x)≡21​sin(2x)cosec(x)≡sin(x)1​⇒cosec(2x)≡sin(2x)1​​​


The integral becomes:

∫1cosec⁡(2x)(cos⁡2(x)+sin⁡2(x)sin⁡(x)cos⁡(x)) dx=∫11sin⁡(2x)(112sin⁡(2x)) dx=∫sin⁡(2x)×2sin⁡(2x) dx=∫2 dx=2x+C\begin{aligned} \int\dfrac{1}{\cosec(2x)}\left( \dfrac{\cos^2(x)+\sin^2(x)}{\sin(x)\cos(x)}\right)\,dx &=\int \dfrac{1}{\frac{1}{\sin(2x)}} \left(\dfrac{1}{\frac12 \sin(2x)}\right) \, dx\\&=\int \sin(2x)\times \dfrac{2}{\sin(2x)}\, dx\\ &=\int 2\, dx\\&=2x+C \end{aligned}∫cosec(2x)1​(sin(x)cos(x)cos2(x)+sin2(x)​)dx​=∫sin(2x)1​1​(21​sin(2x)1​)dx=∫sin(2x)×sin(2x)2​dx=∫2dx=2x+C​​​


​∫1cosec⁡(2x)(cos⁡(x)sin⁡(x)+sin⁡(x)cos⁡(x)) dx=2x+C‾\int \dfrac{1}{\cosec(2x)}\left(\dfrac{\cos(x)}{\sin(x)}+\dfrac{\sin(x)}{\cos(x)}\right)\, dx=\underline{2x+C}∫cosec(2x)1​(sin(x)cos(x)​+cos(x)sin(x)​)dx=2x+C​​​





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Trigonometric identities

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FAQs - Frequently Asked Questions

How many trigonometric identities are there to learn?

There are 15 trigonometric identities to learn in total.

What are the double angle identities?

sin(2x)=2sin(x)cos(x), cos(2x)=cos^2(x)-sin^2(x), tan(2x)=(2tan(x))(1-tan^2(x))

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