Curious Cheetah

# Double u Substitution: ∫tan³θ dθ

## Introduction

$$u$$ substitution is an important technique in integration. In the problem explored in this article, we look at a moderately advanced example in which $$u$$ substitution is performed twice; it also involves trigonometric and logarithmic functions, as well as several other key issues in integration.

## Problem

$\tag{1}\int \tan^3\theta\,d\theta$

## Background Information

In order to complete this problem, we need two related pieces of information: \begin{align} \tag{2a} d(\tan \theta) = \frac{1}{\cos^2 \theta}\,d\theta \\ \tag{2b} \cos^2 \theta = \frac{1}{1+\tan^2 \theta} \end{align}
Note that (2a) is a rephrasing of the standard formula $$d(\tan \theta) = \sec^2 \theta\,d\theta$$. The calculus texts with which Iʼm familiar tend to simply present that and the formulae for $$d(\cot \theta)$$, $$d(\sec \theta)$$, $$d(\csc \theta)$$ as formulae to memorize. However, itʼs important that students do actually work these out; indeed, the derivatives of all six of the main trigonometric functions are based on the simple pair: $$d(\sin \theta) = \cos \theta\,d\theta; d(\cos \theta) = -\sin \theta\,d\theta$$. Memorize that pair, and everything else can be generated as needed.
In this case, we need $$d(\tan \theta)$$. This is derived as follows:
\begin{align} \tag{3a} d(\tan \theta) &= d\left(\frac{\sin \theta}{\cos \theta}\right)\\ \tag{3b} &= \frac{\cos \theta\cdot d(\sin \theta) - \sin \theta\cdot d(\cos \theta)}{\cos^2 \theta}\\ \tag{3c} &= \frac{\cos \theta\cdot\cos \theta\,d\theta - \sin \theta\cdot(-\sin \theta\,d\theta)}{\cos^2\theta}\\ \tag{3d} &= \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\, d\theta \\ \tag{3e} &= \frac{1}{\cos^2 \theta}\, d\theta \end{align}
(3b) follows from the Quotient Rule:$\tag{4} d\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{g(x)^2}$
(3e) follows from the Pythagoran Theorem. Since $$\sin \theta = \frac{a}{c}$$ and $$\cos \theta = \frac{b}{c}$$ (where $$a, b, c$$ are the sides of a right triangle), we can calculate:
\begin{align} \tag{5a} \sin^2 \theta + \cos^2 \theta &= \frac{a^2}{c^2} + \frac{b^2}{c^2} \\ \tag{5b} &= \frac{a^2 + b^2}{c^2} \\ \tag{5c} &= \frac{c^2}{c^2} \\ \tag{5d} &= 1 &c \not= 0 \end{align}
Hence the important identity: $\tag{6} \sin^2 \theta + \cos^2 \theta = 1$
(2b) can be demonstrated as follows: \begin{align} \tag{7a} \tan\theta &= \frac{\sin\theta}{\cos\theta} \\ \tag{7b} \tan^2 \theta &= \frac{\sin^2 \theta}{\cos^2 \theta} & a=\frac{b}{c} &\Rightarrow a^2=\frac{b^2}{c^2} \\ \tag{7c} 1 + \tan^2 \theta &= \frac{\sin^2 \theta}{\cos^2 \theta} + 1 \\ \tag{7d} &= \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} & \cos\theta=0 &\Rightarrow \theta=\pm90^\circ \\ \tag{7e} &= \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} \\ \tag{7f} &= \frac{1}{\cos^2 \theta} \\ \tag{7g} \cos^2 \theta &= \frac{1}{1 + \tan^2 \theta} \end{align}

## Solution

For the first $$u$$ substitution, we have three obvious candidates: $$\tan\theta$$, $$\tan^2\theta$$, and $$\tan^3\theta$$. Letʼs use the first one: \begin{align} \tag{8a} u &= \tan\theta \\ \tag{8b} du &= \frac{d\theta}{\cos^2\theta} \\ \tag{8c} d\theta &= \cos^2\theta \,du \\ \tag{8d} &= \frac{du}{1 + \tan^2\theta} \end{align}
Substituting $$du$$ and then $$u$$ gives us: \begin{align} \tag{9a} \int\tan^3\theta\,d\theta &= \int\frac{\tan^3\theta}{1 + \tan^2\theta}\,du & \text{(8d)} \\ \tag{9b} &= \int\frac{u^3}{1+u^2}\,du & \text{(8a)} \end{align}
We canʼt perform this integration as is, so we have to do another substitution. Letʼs call the second new variable $$v$$. There are several possible substitutions we could try, but letʼs use the denominator: \begin{align} \tag{10a} v &= 1+u^2 \\ \tag{10b} dv &= 2u\,du \\ \tag{10c} du &= \frac{dv}{2u} \end{align}
We substitute $$dv$$ and then $$v$$, and then simplify: \begin{align} \tag{11a} \int \frac{u^3}{1+u^2}\,du &= \int \frac{u^3}{(1+u^2) \cdot 2u}\,dv & \text{(10c)} \\ \tag{11b} &= \int \frac{u^2}{(1+u^2) \cdot 2}\,dv \\ \tag{11c} &= \int \frac{(1+u^2)-1}{2\cdot (1+u^2)}\,dv \\ \tag{11d} &= \int \frac{v - 1}{2v}\,dv & \text{(10a)} \\ \tag{11e} &= \frac{1}{2} \left(\int \frac{v}{v}\,dv - \int \frac{1}{v}\,dv\right) \\ \tag{11f} &= \frac{1}{2} \left(\int 1\,dv - \int \frac{1}{v}\,dv\right) \end{align}
At this point, we can easily integrate (11e): $\tag{12} \frac{1}{2} \left(\int 1\,dv - \int \frac{1}{v}\,dv = \frac{v}{2} - \frac{\ln v}{2}\right) + c$
and then reverse the substitutions: \begin{align} \tag{12a} \frac{1}{2} \left(\int 1\,dv - \int \frac{1}{v}\,dv \right) &= \frac{1}{2}(v - \ln v) + c \\ \tag{12b} &= \frac{v}{2} - \frac{\ln v}{2} + c \end{align}
(7f) (that is, the inverse of (2b)) then gives us: \begin{align} \tag{14a} \frac{1 + \tan^2 \theta}{2} - \frac{\ln(1 + \tan^2\theta)}{2} + c &= \frac{1}{2\cdot\cos^2\theta} - \frac{1}{2}\ln\left(\frac{1}{\cos^2\theta}\right) + c \\ \tag{14b} &= \frac{\sec^2\theta}{2} + \ln((\cos^{-2}\theta)^{-\frac{1}{2}}) + c \\ \tag{14c} &= \frac{\sec^2\theta}{2} + \ln(\cos\theta) + c \end{align}
The second term in (14b) is the result of applying the following property of logarithms: $$a\cdot\log_n c = \log_n c^a$$. Meanwhile, (14c) results from the application of this property of exponents: $$(x^a)^b = x^{(a\cdot b)}$$.
This gives us our final answer (confirmed by Wolfram Alpha):
$\tag{15}\boxed{\int \tan^3\theta\,d\theta = \frac{\sec^2\theta}{2} + \ln(\cos\theta) + c}$