Integration of Trigonometric Expressions

We will introduce an algorithm for obtaining indefinite integrals such as

$\int \frac{(1+\sin(x))}{\sin(x)(1+\cos(x))}\;dx$

or, in general, integral of the form

$\int R(\sin(x), \cos(x))\;dx\quad\quad\quad(1)$

where $R$ is any rational function $R(p, q)$, with $p=\sin(x), q=\cos(x)$.

Let

$t = \tan(\frac{x}{2})\quad\quad(2)$

Solving (2) for $x$, we have

$x = 2\cdot\arctan(t)\quad\quad\quad(3)$

which provides

$\frac{dx}{dt} = \frac{2}{1+t^2}\quad\quad\quad(4)$

and,

$\sin(x) =2\sin(\frac{x}{2})\cos(\frac{x}{2})\overset{\cos^(\frac{x}{2})+\sin^2(\frac{x}{2})=1}{=}\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

yields

$\sin(x) = \frac{2 t}{1+t^2}\quad\quad\quad(5)$

Similarly,

$\cos(x) = \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

gives

$\cos(x)=\frac{1-t^2}{1+t^2}\quad\quad\quad(6)$

We also have (see “Finding Indefinite Integrals” )

$\int f(x)\;dx \overset{x=\phi(t)}{=} \int f(\phi(t))\cdot\frac{d\phi(t)}{dt}\;dt$.

Hence

$\int R(\cos(x), \sin(x))\;dx \overset{(2), (4), (5), (6)}{=} \int R(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})\cdot\frac{2}{1+t^2}\;dt$,

and (1) is reduced to an integral of rational functions in $t$.

Example-1 Evaluate $\int \csc(x)\;dx$.

Solution: $\csc(x) = \frac{1}{\sin(x)}\implies \int \csc(x)\;dx = \int \frac{1}{\sin(x)}\;dx$

$= \int \frac{1}{\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\;dt=\int\frac{1}{t}\;dt = \log(t) = \log(\tan(\frac{x}{2}))$.

Example-2 Evaluate $\int \sec(x)\;dx$.

Solution: $\sec(x) = \frac{1}{\cos(x)}\implies \int \sec(x)\; dx =\int \frac{1}{\cos(x)}\;dx$

$= \int \frac{1}{\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\; dt=\int \frac{2}{1-t^2}\;dt=\int \frac{2}{(1+t)(1-t)}\;dt=\int \frac{1}{1+t} + \frac{1}{1-t}\;dt$

$=\int \frac{1}{1+t}\;dt - \int \frac{-1}{1-t}\;dt$

$=\log(1+t) -\log(1-t) =\log\frac{1+t}{1-t}=\log(\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})})$.

According to CAS (see Fig. 1),

Fig. 1

However, the two results are equivalent as a CAS-aided verification (see Fig. 2) confirms their difference is a constant (see Corollary 2 in “Sprint to FTC“).

Fig. 2

Exercise-1 According to CAS,

Show that it is equivalent to the result obtained in Example-1

Exercise-2 Try

$\int \frac{1}{\sin(x)+1}\;dx$

$\int \frac{1}{\sin(x)+\cos(x)}\;dx$

$\int \frac{1}{(2+\cos(x))\sin(x)}\;dx$

$\int \frac{1}{5+4\sin(x)}\;dx$

$\int \frac{1}{2\sin(x)-\cos(x)+5}\;dx$

and of course,

$\int \frac{1+\sin(x)}{\sin(x)(1+\cos(x))}\;dx$