# Standard Integrals

There are some integrals it's helpful to keep in mind so that when we come across a tricky problem, we know what certain integrals equal, and we can have less work to do. In addition, the methods that are used to solve these integrals will apply to similar integrals, so it is worth knowing these.

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## Trigonometric integrals

Not all trigonometric integrals are easy to solve, even for basic trigonometric functions. As such, we will examine the integrals of some base trigonometric functions.

### 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}

Remember to add the integration constant at the end.

### 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.

## Standard Integrals - Key takeaways

It's helpful to keep some common integrals in mind. Here's the list of standard integrals:

• $\int \csc (ax)\, \mathrm{d} x = -\frac{1}{a}\ln\left|\csc(ax)+\cot(ax) \right| + C$
• $\int\sec(ax)\, \mathrm{d}x = -\frac{1}{a}\ln\left|\sec(ax)+\tan(ax) \right| + C$
• $\int\tan(ax)\, \mathrm{d}x = \frac{1}{a}\ln\left|\sec(ax)\right| + C$
• $\int\cot(ax)\, \mathrm{d}x =\frac{1}{a} \ln\left|\sin(ax)\right| + C$
• $\int \frac{1}{x^2+a^2}\, \mathrm{d} x = \frac{1}{a}\arctan \left(\frac{x}{a}\right) + C$
• $\int \frac{1}{\sqrt{a^2-x^2}}\, \mathrm{d} x = arcsin \left(\frac{x}{a}\right) + D$
• $\int \frac{1}{x^2-a^2} \, \mathrm{d} x = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C$
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What are standard integrals?

A standard integral is one that behaves a bit differently, and/or comes up a lot, and it is, therefore, useful to keep the formula in mind.

How do you verify a standard integral?

Like any integral, we can differentiate the result to ensure we get back to what we integrated.

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