## 17Calculus Integrals - Tangent Reduction Formula

This page covers the derivation and use of the tangent reduction formula for integration.

Tangent Reduction Formula (n is an integer and $$n>1$$)

$$\displaystyle{\int{\tan^n x~dx}= \frac{\tan^{n-1}x}{n-1} - \int{\tan^{n-2}x~dx}}$$

When you have an integral with only tangent where the power is greater than one, you can use the tangent reduction formula, repeatedly if necessary, to reduce the power until you end up with either $$\tan x$$ or $$\tan^2 x$$. Let's derive the formula and then work some practice problems.

Deriving The Tangent Reduction Formula

$$\displaystyle{ \int{\tan^nx~dx} = \frac{\tan^{n-1}x}{n-1} - \int{\tan^{n-2}x~dx} }$$ Separate out a $$\tan^2x$$. $$\int{\tan^nx~dx} = \int{\tan^{n-2}x \tan^2x~dx}$$ Use the identity $$\sec^2x=1+\tan^2x$$ to replace the $$\tan^2x$$ with $$\sec^2x-1$$. $$\int{\tan^nx~dx} = \int{\tan^{n-2}x~(\sec^2x-1)~dx}$$ Distribute the $$\tan^{n-2}x$$ term and separate into two integrals. $$\int{\tan^nx~dx} =$$ $$\int{\tan^{n-2}x \sec^2x~dx}$$ $$-$$ $$\int{\tan^{n-2}x~dx}$$ Use integration by substitution on the first integral on the right side of the equal sign. $$u=\tan x \to du=\sec^2x~dx$$ $$\int{\tan^nx~dx} =$$ $$\int{u^{n-2}~du}$$ $$-$$ $$\int{\tan^{n-2}x~dx}$$ Integrate the term containing u and convert back to x's. $$\displaystyle{ \int{\tan^nx~dx} = }$$ $$\displaystyle{ \frac{u^{n-1}}{n-1} - \int{\tan^{n-2}x~dx} }$$ This last equation is the tangent reduction formula.

Now let's work some practice problems.

Practice

Unless otherwise instructed, evaluate these integrals directly, then check your answer using the reduction formula.

$$\displaystyle{ \int{ \tan^3x ~dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int{ \tan^3x ~dx } }$$ using the techniques on the tangent-secant trig integration page. Then check your answer using the tangent reduction formula.

Solution

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$$\displaystyle{ \int{ \tan^4x ~dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int{ \tan^4x ~dx } }$$ using the techniques on the tangent-secant trig integration page. Then check your answer using the tangent reduction formula.

Solution

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Evaluate $$\int{ \tan^5x ~dx }$$ using the tangent reduction formula.

Problem Statement

Evaluate $$\int{ \tan^5x ~dx }$$ using the tangent reduction formula.

Solution

Here are two videos, by two different instructors, solving this problem.

### 116 video

video by PatrickJMT

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### trig integration 17calculus youtube playlist

You CAN Ace Calculus

 basic trig integration integration by substitution integration by parts

### Trig Identities and Formulas

basic trig identities

$$\sin^2\theta+\cos^2\theta=1$$   |   $$1+\tan^2\theta=\sec^2\theta$$

$$\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}$$   |   $$\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}$$

$$\displaystyle{\sec\theta=\frac{1}{\cos\theta}}$$   |   $$\displaystyle{\csc\theta=\frac{1}{\sin\theta}}$$

power reduction (half-angle) formulae

$$\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}$$   |   $$\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}$$

double angle formulae

$$\sin(2\theta)=2\sin\theta\cos\theta$$   |   $$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$

list of trigonometric identities - wikipedia

trig sheets - pauls online notes

17calculus trig formulas - full list

### Trig Derivatives and Integrals

basic trig derivatives

$$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$

$$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$

$$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$

$$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$

$$\displaystyle{ \frac{d[\sec(t)]}{dt} = }$$ $$\sec(t)\tan(t)$$

$$\displaystyle{ \frac{d[\csc(t)]}{dt} = }$$ $$-\csc(t)\cot(t)$$

basic trig integrals

$$\int{\sin(x)~dx} = -\cos(x)+C$$

$$\int{\cos(x)~dx} = \sin(x)+C$$

$$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$

$$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$

$$\int{\sec(x)~dx} = \ln\abs{\sec(x)+\tan(x)}+C$$

$$\int{\csc(x)~dx} = -\ln\abs{\csc(x)+\cot(x)}+C$$

reduction formulae

Reduction Formulas (n is a positive integer)

$$\displaystyle{\int{\sin^n x~dx} = -\frac{\sin^{n-1}x\cos x}{n} + }$$ $$\displaystyle{ \frac{n-1}{n}\int{\sin^{n-2}x~dx} }$$

$$\displaystyle{\int{\cos^n x~dx} = \frac{\cos^{n-1}x\sin x}{n} + }$$ $$\displaystyle{ \frac{n-1}{n}\int{\cos^{n-2}x~dx}}$$

Reduction Formulas (n is an integer and $$n>1$$)

$$\displaystyle{\int{\tan^n x~dx}= \frac{\tan^{n-1}x}{n-1} - \int{\tan^{n-2}x~dx}}$$

$$\displaystyle{\int{\sec^n x~dx} = \frac{\sec^{n-2}x\tan x}{n-1} + }$$ $$\displaystyle{ \frac{n-2}{n-1}\int{\sec^{n-2}x~dx}}$$

17calculus trig formulas - full list

related topics on other pages

basic trig integration

secant-tangent trig integration

Wikipedia - List of Trig Identities

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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