Standard Integrals

The integral of cosecant

Suppose we wish to integrate \(I = \int \csc (ax)\, \mathrm{d} x\), with \(a\) being the constant.

This, at first glance, seems pretty intimidating, but this is an occasion where we can use a trick.

Let us multiply the integral by

\[ \frac{\csc (ax) + \cot(ax)}{\csc(ax)+\cot(ax)}.\]

We can do this, as it is equivalent to multiplying by 1.

This gives

\[ \int \csc (ax)\left(\frac{\csc (ax) + \cot(ax)}{\csc(ax)+\cot(ax)} \right)\, \mathrm{d} x. \]

Then, let's use the substitution of \(u= \csc (ax) + \cot(ax) \), meaning that

\[ \begin{align} \frac{\mathrm{d}u}{\mathrm{d}x} &= -a\csc (ax) \cot(ax) - a\csc^2 (ax) \\ &= -a\csc(ax)(\cot(ax)+\csc(ax)), \end{align}\]

and

\[ \mathrm{d}x = \frac{\mathrm{d}u }{-a\csc(ax)(\cot(ax)+\csc(ax)) }.\]

Substituting this in gives

\[ \begin{align} \int \csc (ax)\left(\frac{\csc (ax) + \cot(ax)}{\csc(ax)+\cot(ax)} \right)\, \mathrm{d} x &= \frac{\csc(ax)(\cot(ax)+\csc(ax)) } {\cot(ax)+\csc(ax) }\, \mathrm{d} x \\ &= -\frac{1}{a}\int \frac{1}{u} \, \mathrm{d} x . \end{align}\]

This is now easy to evaluate. This gives

\[ \begin{align} I &= -\frac{1}{a}\ln|u| \\ &= -\frac{1}{a}\ln\left|\csc(ax)+\cot(ax) \right| + C. \end{align}\]

The integral of secant

This is similar to the previous example.

Define \(J = \int\sec(ax)\, \mathrm{d}x\), with \(a\) as the constant.

This time, we will multiply the integral by

\[ \frac{\sec (ax) + \tan(ax)}{\sec(ax)+\tan(ax)}.\]

This gives

\[ J = \int \frac{\sec(ax)(\sec (ax) + \tan(ax) )}{\sec (ax) + \tan(ax) }\, \mathrm{d}x .\]

Now, let \(u = \sec (ax) + \tan(ax) \), which will give

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = a\sec (ax) \tan(ax) +a\sec^2 (ax) . \]

This implies that

\[ \mathrm{d}x = \frac{\mathrm{d}u }{a\sec (ax) \tan(ax) + a\sec^2 (ax) }.\]

Filling this in, we get that

\[ J = -\frac{1}{a}\ln|u| .\]

We can now evaluate this to get

\[ \begin{align} J &= -\frac{1}{a}\ln|u| \\ &= -\frac{1}{a}\ln\left|\sec(ax)+\tan(ax) \right| + C. \end{align}\]

where again we have added the integration constant at the end.

The integral of tangent

This requires a different approach and does not require any additional tricks to work it out.

Define \(K = \int\tan(ax)\, \mathrm{d}x\), with \(a\) as the constant.

Recall that the definition of

\[ \tan(ax) = \frac{\sin(ax)}{\cos(ax)},\]

and then we can write

\[ K = \int\frac{\sin(ax)}{\cos(ax)} \, \mathrm{d}x .\]

We can now use a substitution of \(u = \cos (ax)\), and so

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = -a\sin(ax).\]

This means that

\[ \mathrm{d}x = \frac{\mathrm{d}u }{-a\sin(ax) }.\]

Using the laws of logs, we can tidy this up to give

\[ K = \frac{1}{a}\ln\left|\sec(ax)\right| + C.\]

The integral of cotangent

Define \(L = \int\cot(ax)\, \mathrm{d}x\), with \(a\) as the constant. Recall that the definition of

\[ \cot(ax) = \frac{\cos(ax)}{\sin(ax)},\]

and then we can write

\[ L = \int\frac{\cos(ax)}{\sin(ax)} \, \mathrm{d}x .\]

Now make the substitution of \(u = \sin (ax)\). This implies that

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = a\cos(ax).\]

Filling this in, we obtain that

\[ \begin{align} L &= \frac{1}{a} \int\frac{1}{u} \, \mathrm{d}x \\ &= \frac{1}{a}\ln|u| + C \\ &= \frac{1}{a} \ln\left|\sin(ax)\right| + C. \end{align}\]

Useful reciprocal polynomial integrals

Usually, when we integrate polynomials, there is often a simple integral. However, in these examples, the answers appear to come out of nowhere.

Integral of \( \dfrac{1}{x^2+a^2}\)

Define

\[ I = \int \frac{1}{x^2+a^2}\, \mathrm{d} x.\]

Now we are going to substitute \(x = a\tan(t)\). This means that

\[ \frac{\mathrm{d}x}{\mathrm{d}t} = a\sec^2 (at),\]

and so \( \mathrm{d}x = a\sec^2 (at) \mathrm{d}t \). Filling this in, we get

\[ I = \int \frac{a\sec^2(t)}{a^2(\tan^2(t)+1)}\, \mathrm{d} t.\]

We can now use the trigonometric identity of

\[1 + \tan^2(t) = \sec^2(t) \]

to give

\[ \begin{align} I &= \int \frac{a\sec^2(t)}{a^2(\tan^2(t)+1)}\, \mathrm{d} t \\ &= \frac{1}{a}\int 1 \, \mathrm{d} t .\end{align}\]

This integral is now trivial, giving

\[ I = \frac{1}{a}t + C.\]

Filling in for \(t\), we find

\[ I = \frac{1}{a}\arctan \left(\frac{x}{a}\right) + C.\]

Integral of \( \dfrac{1}{x^2-a^2}\)

This integral calls for partial fractions. We first factorise

\[ x^2-a^2 = (x+a)(x-a).\] This means we look to split

\[ \frac{1}{x^2-a^2 } = \frac{A}{x-a} + \frac{B}{x+a}.\]

Multiplying through by \(x^2-a^2\) we obtain

\[ \begin{align} 1 &= A(x+a)+B(x-a) \\ &= (A+B)x + (A-B)a. \end{align}\]

This implies that \(A + B = 0\) and

\[ A - B = \frac{1}{a},\] which further implies

\[ A = \frac{1}{2a} \text{ and } B = -\frac{1}{2a}.\]

This will give that

\[ \begin{align}\int \frac{1}{x^2-a^2} \, \mathrm{d} x &= \frac{1}{2a} \int \frac{1}{x-a} \, \mathrm{d} x - \frac{1}{2a}\int \frac{1}{x+a} \, \mathrm{d} x \\ &= \frac{1}{2a} \left( \ln|x-a| - \ln|x+a|\right) + C. \end{align}\]

We can further simplify this using law of logs, which yields

\[ \int \frac{1}{x^2-a^2} \, \mathrm{d} x = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C.\]

Integral of \( \dfrac{1}{\sqrt{a^2-x^2}}\)

For this integral, we have two methods, which yield results that appear different but are equivalent. Define

\[ J = \int \frac{1}{\sqrt{a^2-x^2}}\, \mathrm{d} x.\]

Method 1:

Use the substitution \(x = a\cos (t)\). Then

\[ \frac{\mathrm{d}x}{\mathrm{d}t} = -a\sin t,\]

which gives \( \mathrm{d}x = -a\sin t \mathrm{d}t \). Filling this in, we obtain

\[ \begin{align} J &= \int \frac{-a\sin t}{\sqrt{a^2-a^2\cos^2 t}}\, \mathrm{d} t \\ &= \int \frac{-a\sin t}{a\sqrt{1- \cos^2 t}}\, \mathrm{d} t . \end{align}\]

We are now able to use the trigonometric identity of

\[ \sin^2 t + \cos^2 t = 1,\]

or rearranged

\[ 1 - \cos^2 t = \sin^2 t,\]

to give

\[ \begin{align} J &= \int \frac{-a\sin t}{a\sin t}\, \mathrm{d} t \\ &= \int -1\, \mathrm{d} t \\ &= -t+C \\ &= -\arccos \left(\frac{x}{a}\right) + C. \end{align}\]

Method 2:

Use the substitution \(x = a\sin t \). Then

\[ \frac{\mathrm{d}x}{\mathrm{d}t} = a\cos t,\]

which gives \( \mathrm{d}x = a\cos t \mathrm{d}t \). Filling this in, we obtain

\[ \begin{align} J &= \int \frac{-a\cos t}{\sqrt{a^2-a^2\sin^2 t}}\, \mathrm{d} t \\ &= \int \frac{-a\cos t}{a\sqrt{1- \sin^2 t}}\, \mathrm{d} t . \end{align}\]

We are now able to use the trigonometric identity of

\[ 1 - \sin^2 t = \cos^2 t,\]

to give

\[ \begin{align} J &= \int \frac{a\cos t}{a\cos t}\, \mathrm{d} t \\ &= \int 1\, \mathrm{d} t \\ &= t+C \\ &= \arcsin \left(\frac{x}{a}\right) + D. \end{align}\]

Why do these two match up? We can see that

\[ -\frac{\mathrm{d}}{\mathrm{d} x} \arccos \left(\frac{x}{a}\right) = \frac{\mathrm{d}}{\mathrm{d} x} \arcsin \left(\frac{x}{a}\right) \]

so these two should be equal, although the constants are not identical, but are related to one another in some way.