Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to solve $\int\tan^3(x)\,dx$ ? Do I use substitution?

share|improve this question
1  
Wikipedia has a nice reference page for all of these: en.wikipedia.org/wiki/… –  Eric Naslund Feb 2 '11 at 18:15
    
Thanks Eric, that is so useful! –  xiamx Feb 2 '11 at 18:34

3 Answers 3

up vote 3 down vote accepted

One standard approach to integral involving powers of tangent is the following: Rewrite your power as something times $\tan^2(x)$. In this case: $\tan^3(x)=\tan(x)\cdot\tan^2(x)$.

Now, use that $\tan^2(x)=\sec^2(x)-1$, so $$ \int\tan^3(x)dx=\int\tan(x)\sec^2(x)dx-\int\tan(x)dx.$$ The first integral on the right hand side can be solved using the substitution $u=\tan(x)$, so $du=\sec^2(x)dx$ and $$ \int\tan(x)\sec^2(x)dx=\int u du. $$ The second integral is $\ln|\sec x|$, and can be found writing $\tan(x)=\sin(x)/\cos(x)$ and using the substitution $v=\cos(x)$.


Note that the same idea allows you to solve $\displaystyle \int\tan^n(x) dx$ for any $n\ge 3$. For example: $$\int\tan^5(x)dx=\int\tan^3(x)\tan^2(x)dx=\int\tan^3(x)(\sec^2(x)-1)dx,$$ as before we split this last integral into two: $$\int\tan^3(x)\sec^2(x)dx-\int\tan^3(x)dx.$$ The first one is solved with the substitution $u=\tan(x)$. The second is the integral we solved before.

This, by the way, is an example of a reduction method that transforms an integral that depends on an integer parameter $n$ (in this case, $\displaystyle \int\tan^n(x)dx$) into a similar integral, but that now depends on some $m<n$. This gives us an algorithm to solve the integral by repeatedly applying the reduction procedure.

share|improve this answer
    
@Sivaram: It works for all $n$: For a typical example for even $n$: $\int\tan^4(x)dx=\int\tan^2(x)\sec^2(x)dx-\int \tan^2(x)dx=\int\tan^2(x)\sec^2(x)dx-\int\sec^2(x)dx+\int dx$. The last integral is just $x+C$, and for the other two, the substitution $u=\tan(x)$ works. –  Andres Caicedo Feb 2 '11 at 16:32
    
Oh sorry. Sorry for comment. I withdraw the previous comment. –  user17762 Feb 2 '11 at 16:36

A simple way out is to write $\tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}$. Now let $\cos(x) = t$ and rewrite $\sin^2(x) = 1-t^2$.

The integral now becomes $\int \frac{\sin^3(x)}{\cos^3(x)} dx$.

$\cos(x) = t \Rightarrow -\sin(x)dx = dt$ and $\sin^2(x) = 1-t^2$.

Note the numerator of the integral $\sin^3(x) dx$ can be written as $\sin^2(x) \times \sin(x) dx = (1-t^2) (-dt) = (t^2-1) dt$

So the integral becomes $\int \frac{t^2-1}{t^3} dt = \int \frac{dt}{t} - \int \frac{dt}{t^3} = \log(t) + \frac{1}{2t^2} = \log(\cos(x)) + \frac{1}{2 \cos^2(x)} = \log(\cos(x)) + \frac{\sec^2(x)}{2}$

share|improve this answer
    
I like this method a lot! –  frogeyedpeas Jul 17 at 18:21

HINT: Rewrite your integral in this form:

\begin{align*} \int\tan^{3}{x} \ dx & = \int (\sec^{2}{x}-1) \cdot \tan{x} \ dx \\ &= \int \sec^{2}{x} \cdot \tan{x} \ dx - \int \tan{x} \ dx \\ &= \int \sec{x} \cdot (\sec{x}\tan{x}) \ dx - \int \tan{x} \ dx \end{align*}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.