How could we integrate $\tan^3 \theta$?

How could we integrate $\tan^3 \theta$?

Question

1. Evaluate

$$\begin{equation*} \int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta, \end{equation*}$$

giving the answer correct to two significant figures.

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1. By using the substitution $x+1 = u^2$, or otherwise, find

$$\begin{equation*} \int \frac{\sqrt{x+1}}{x-3} \:dx. \end{equation*}$$

Solution

We can use the identity

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

to help us:

$$\begin{align*} \int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta &= \int_0^{\tfrac{1}{4} \pi} (\tan^2 \theta) \tan \theta \:d\theta\\ &= \int_0^{\tfrac{1}{4} \pi} (\sec^2 \theta-1) \tan \theta \:d\theta\\ &= \int_0^{\tfrac{1}{4} \pi} \sec^2 \theta\tan \theta - \tan \theta \:d\theta\\ \end{align*}$$

So we have two integrals to calculate; the first is

$$\int_0^{\tfrac{1}{4} \pi} \sec^2 \theta\tan \theta \:d\theta$$

We notice that $\dfrac{d}{d \theta} \tan \theta = \sec^2 \theta$ (this is a standard result). So, letting $u = \tan \theta$, our integral becomes

$$\int_0^{\tfrac{1}{4} \pi} u \dfrac{du}{d\theta} \:d\theta = \int_0^{\tan \tfrac{\pi}{4}} u \:du = \dfrac{1}{2} \left[\dfrac{u^2}{2}\right]_0^{\tan \tfrac{\pi}{4}} = \dfrac{1}{2} - 0 = \dfrac{1}{2}$$

Our second integral is

$$\int_0^{\tfrac{1}{4} \pi} tan \theta \:d\theta = \int_0^{\tfrac{1}{4} \pi} \dfrac{\sin \theta}{\cos \theta} \:d\theta$$

We can calculate this by writing $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$, substituting $v = \cos \theta$, and noticing that $-\sin \theta \:d\theta = \:dv$, so our integral is in fact

$$\int_0^{\tfrac{1}{4} \pi} \dfrac{-1}{v} \dfrac{\:d v}{\:d \theta} \:d\theta = \int_0^{\cos \tfrac{\pi}{4}} \dfrac{-1}{v} \:dv = - \biggl[ \ln v \biggr]_1^{\cos \tfrac{\pi}{4}} = \biggl[ \ln \dfrac{1}{v} \biggr]_1^{\tfrac{1}{\sqrt 2}} = \ln{\sqrt 2} - \ln 1 = \dfrac{1}{2} \ln 2$$

So our final value for $\int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta$ is

$$\dfrac{1}{2} - \dfrac{1}{2} \ln 2 = 0.1534...$$

which equals

$0.15$

to two significant figures.

Alternatively, we can use substitution in a different way. Let

$u = \tan \theta$

, so that

$$\begin{equation*} \frac{du}{d\theta} = \sec^2 \theta = 1 + \tan^2 \theta = 1 + u^2 \end{equation*}$$

and hence

$$\begin{align*} \int_0^{\tfrac{1}{4} \pi} \tan^3 \theta \:d\theta = \int_0^1 \frac{u^3}{1 + u^2} \:du &= \int_0^1 \frac{u(1 + u^2)}{1 + u^2} - \frac{u}{1 + u^2} \:du \\ &= \left[ \frac{u^2}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \frac{2u}{1 + u^2} \:du \\ &= \frac{1}{2} - \frac{1}{2} \left[ \log(1 + u^2) \right]_0^1 \\ &= \frac{1}{2} - \frac{\log 2}{2} \\ &= 0.1534\!\ldots \end{align*}$$

which equals $0.15$ to two significant figures.

Take $x+1 = u^2$, so that $x = u^2 - 1$, and $dx = 2u du$.

Hence,

$$\begin{align*} \int \frac{\sqrt{x+1}}{x-3} \:dx &= \int \frac{u}{u^2-4} 2u\:du \\ &= \int \frac{2u^2 - 8+8}{u^2 - 4} \:du\\ &= \int 2 + \frac{8}{(u-2)(u+2)} \:du. \end{align*}$$

We can use partial fractions,

$$\begin{align*} \frac{8}{(u-2)(u+2)} &= \frac{A}{u-2} + \frac{B}{u+2}\\ \implies\quad 8 &= A(u+2) + B(u-2) \end{align*}$$

and substituting

$u=\pm 2$

we find

$A=2$

and

$B=-2$

. Hence,

$$\begin{align*} \int \frac{\sqrt{x+1}}{x-3} \:dx &= 2u + 2 \int \frac{1}{(u-2)} - \frac{1}{(u+2)} \:du \\ &= 2u + 2 \log(u-2) - 2\log(u+2) + c \\ &= 2\sqrt{x+1} + 2\log(\sqrt{x+1} - 2) - 2\log(\sqrt{x+1} + 2) + c. \end{align*}$$