175 – Integrating products of secants and tangents

(In what follows, I will write {\tan^n x} for {(\tan x)^n} and {\sec^m x} for {(\sec x)^m.})

Recall that

\displaystyle \tan x=\frac{\sin x}{\cos x}\quad\mbox{ and }\quad\sec x=\frac1{\cos x},

that

\displaystyle (\tan x)'=\sec^2x\quad\mbox{ and }\quad(\sec x)'=\sec x\cdot\tan x,

and that

\displaystyle \sec^2x=\tan^2x+1.

The formulas below make use of these identities repeatedly.

We want a series of methods and reduction formulas that allow us to evaluate any expression of the form

\displaystyle \int \sec^m x\cdot \tan^n x\,dx,

for {m} and {n} integers, {m,n\ge0.}

 

1. Integrating powers of {\sec x}

In lecture we saw that

\displaystyle \int\sec x\,dx=\ln|\sec x+\tan x| +C,

and derived the following reduction formula for integrating powers of {\sec x:}

\displaystyle \int\sec^m x\,dx=\frac{\sec^{m-2}x\cdot\tan x}{m-1}+\frac{m-2}{m-1}\int\sec^{m-2}x\,dx +C.

This formula is valid for any {m\ge2.}

(The integral of {\sec x} was obtained by what is essentially the method of partial fractions, that we will study in detail when we reach section 7.4 in the book. The reduction formula was obtained using integration by parts.)

Exercise 1 Find {\displaystyle \int\sec^4 x\,dx} and {\displaystyle \int\sec^5 x\,dx.}

 

2. Integrating powers of {\tan x}

 

We also saw that

\displaystyle \int\tan x\,dx=\ln|\sec x|+C,

and derived the following reduction formula:

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

valid for all {n\ge2.}

Exercise 2 Find {\displaystyle \int\tan^4 x\,dx} and {\displaystyle \int\tan^5 x\,dx.}

 

3. Integrating products of powers of {\sec x} and {\tan x}

 

Suppose now that we need to evaluate an expression of the form

\displaystyle \int \sec^mx\cdot\tan^nx\,dx,

where both {n} and {m} are at least 1. As in the case of integrals of products of powers of sines and cosines, it is best to divide the problem into two cases.

 

3.1. If {n} is odd

 

Suppose first that {n} is odd, say {n=2k+1,} for some integer {k\ge0.} Then

\displaystyle \int \sec^mx\cdot\tan^nx\,dx=\int \sec^{m-1}x\cdot\tan^{2k}x\cdot(\sec x\cdot\tan x)\,dx.

Since {\tan^2x=\sec^2x-1,} we can write {\tan^{2k}x=(\sec^2x-1)^k.} The last integral can then be expressed in the form

\displaystyle \int \sec^{m-1}x\cdot(\sec^2 x-1)^k\cdot(\sec x\cdot\tan x)\,dx.

This can be easily evaluated using the substitution {u=\sec x,} that transforms it into

\displaystyle \int u^{m-1} (u^2-1)^k\,du.

To evaluate this expression, expand {(u^2-1)^k} and multiply the result by {u^{m-1}.} This gives us a polynomial in {u.} We can integrate the polynomial term by term, and then replace {\sec x} back in place of {u.}

Exercise 3 Find {\displaystyle \int \sec^3x\cdot\tan x\,dx} and {\displaystyle \int \sec^4x\cdot\tan^5 x\,dx.}

 

3.2. If {n} is even

 

Suppose now that {n} is even, say {n=2k} for some integer {k\ge1.} Then

\displaystyle \int\sec^m x\cdot\tan^n x\,dx=\int\sec^m x\,(\sec^2x-1)^k\,dx.

To evaluate this expression, expand {(\sec^2x-1)^k} and multiply the result by {\sec^m x.} This gives us a sum of powers of {\sec x,} that can be evaluated term by term using the reduction formula from Section 1. Note that this method works even if {m=0.}

Exercise 4 Find {\displaystyle \int \sec^3x\cdot\tan^4\,dx.}

 

Exercise 5 Find {\displaystyle \int \tan^4 x\,dx} using this method, and show that your answer actually gives the same result as the answer you found in Exercise 2.

 

4. Integrating powers of {\csc x} and {\cot x}

 

Exercise 6 (This is long.) Explain how to adapt the methods from the previous sections to find any integral of the form {\displaystyle \int \csc^m x\cdot\cot^n x\,dx.} Then repeat the previous exercises but with {\csc} in place of {\sec,} and {\cot} instead of {\tan.}

 

Although I am not writing this as an exercise, it is a good idea to also spend some time thinking about what one would do with integrals of products of powers of {\sec x} and {\cot x,} or of {\csc x} and {\tan x,} or of all four expressions.

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