Integration using Inverse Trigonometric Functions

Here we will take a look at the derivative of the inverse sine function and how we can use it in integration.

$$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{\frac{x}{a}}=\frac{1}{\sqrt{a^2-x^2}}$$

Let's now consider the following integral:

$$\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}$$

The above integral can be done using Trigonometric Substitution. Here we would need to do \(u=a\sin{x}\) and use the Pythagorean Trigonometric Identity \(\sin^2{x}+\cos^2{x}=1\). Instead, we will use our refresher on the derivative of the inverse sine function.

$$\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\arcsin{\frac{x}{a}}+C$$

This becomes way easier as long as we have an Integration Table on hand!

It is the turn of the derivative of the inverse cosine function.

$$\frac{\mathrm{d}}{\mathrm{d}x}\arccos{\frac{x}{a}}=\text{-}\frac{1}{\sqrt{a^2-x^2}}$$

Notice something? This is the negative of the derivative of the inverse sine function. Consider now the following integral.

$$\int\text{-}\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}$$

We can use the inverse cosine function, but we can also factor the negative sign out of the integral and use the inverse sine function.

$$\int\text{-}\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\text{-}\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}$$

This way, there is one less formula we need to memorize.

$$\int\text{-}\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\text{-}\arcsin{\frac{x}{a}}+C$$

Evaluate the following integral:

$$\int\frac{\mathrm{d}x}{9+x^2}$$

When evaluating these types of integrals, we need to identify the \(a\) value; this way, we can use the corresponding formula to find the antiderivative of the requested function.

• Rewrite \(9\) as \(3^2\).

$$\int\frac{\mathrm{d}x}{9+x^2}=\int\frac{\mathrm{d}x}{3^2+x^2}$$

• Integrate using the inverse tangent function with \(a=3\).

$$\int\frac{\mathrm{d}x}{9+x^2}=\frac{1}{3}\arctan{\frac{x}{3}}+C$$

Evaluate the following integral:

$$\int\frac{\mathrm{d}x}{\sqrt{5-x^2}}$$

Even though \(5\) is not a perfect square, it is still the square of \(\sqrt{5}\).

• Rewrite \(5\) as \( (\sqrt{5})^2 \).

$$\int\frac{\mathrm{d}x}{\sqrt{5-x^2}}=\int\frac{\mathrm{d}x}{\sqrt{(\sqrt{5})^2-x^2}}$$

• Integrate using the inverse sine function with \(a=\sqrt{5}\).

$$\int\frac{\mathrm{d}x}{\sqrt{5-x^2}}=\arcsin{\frac{x}{\sqrt{5}}}+C$$

Evaluate the following integral:

$$\int\frac{x\,\mathrm{d}x}{\sqrt{1-x^4}}$$

Since the variable is being subtracted, we will use the inverse sine function. Next, we note that we can rewrite \(x^4=(x^2)^2\).

$$\int\frac{x\,\mathrm{d}x}{\sqrt{1-x^4}}=\int\frac{x\,\mathrm{d}x}{\sqrt{1-(x^2)^2}}$$

This suggests using Integration by Substitution with \(u=x^2\).

• Use The Power Rule to find \(\mathrm{d}u\).

$$\frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \frac{1}{2}\mathrm{d}u=x\,\mathrm{d}x$$

• Write the integral in terms of \(u\) and \(\mathrm{d}u\).

$$\int\frac{x\,\mathrm{d}x}{\sqrt{1-x^4}}=\frac{1}{2}\int\frac{\mathrm{d}u}{\sqrt{1-u^2}}$$

• Integrate using the inverse sine function.

$$\int\frac{x\,\mathrm{d}x}{\sqrt{1-x^4}}=\frac{1}{2}\arcsin{u}+C$$

• Substitute back \(u=x^2\).

$$\int\frac{x\,\mathrm{d}x}{\sqrt{1-x^4}}=\frac{1}{2}\arcsin{x^2}+C$$

Evaluate the following integral:

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}$$

Begin by noting that \(x^6=(x^3)^2\).

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\int\frac{\mathrm{d}x}{x\sqrt{(x^3)^2-4}}$$

This time we will use the inverse secant function since a constant is being subtracted from the variable, but first we need to make the following substitution:

$$u=x^3 \rightarrow \mathrm{d}u=3x^2\,\mathrm{d}x$$

We need to have \(3x^2\) in the numerator, so this substitution can work. Hence we will multiply by 1 inside the integral.

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\frac{1}{3}\int\frac{3x^2\,\mathrm{d}x}{x^2\,x\sqrt{(x^3)^2-4}}$$

Here, we have multiplied by \(3x^2\) in the numerator and the denominator and factored \(\frac{1}{3}\) out of the integral. This expression can now be simplified as follows:

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\frac{1}{3}\int\frac{3x^2\,\mathrm{d}x}{x^3\sqrt{(x^3)^2-4}}$$

Substituting \(u=x^3\) we get:

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\frac{1}{3}\int\frac{\mathrm{d}u}{u\sqrt{u^2-4}}$$

We can now integrate using the inverse secant function.

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\frac{1}{3}\cdot\frac{1}{2}\mathrm{arcsec}{\frac{u}{2}}+C$$

Finally, we substitute back \(u=x^3\). We also simplify the fractions.

$$\int\frac{\mathrm{d}x}{x\sqrt{x^6-4}}=\frac{1}{6}\mathrm{arcsec}{\frac{x^3}{2}}+C$$

Evaluate the following integral:

$$\int \mathrm{arctan}{\, 3x} \, \mathrm{d}x.$$

This integral almost looks like the integral of just the inverse tangent function. We just need to do a substitution to solve it.

• Let \( u=3x.\) Find \( \mathrm{d}u \) by using The Power Rule.

$$u=3x \rightarrow \mathrm{d}u=3\mathrm{d}x$$

• Isolate \( \mathrm{d}x. \)

$$\mathrm{d}u=3\mathrm{d}x \rightarrow \mathrm{d}x=\frac{1}{3}\mathrm{d}u$$

• Write the integral in terms of \(u\) and \(\mathrm{d}u.\)

$$\begin{align} \int \mathrm{arctan}{\, 3x} \, \mathrm{d}x &= \int \mathrm{arctan}{\, u}\left( \frac{1}{3} \mathrm{d}u \right) \\ &= \frac{1}{3}\int \mathrm{arctan}{\, u} \, \mathrm{d} u \end{align}$$

• Integrate the inverse tangent function.

$$\int \mathrm{arctan}{\, 3x} \, \mathrm{d}x = \frac{1}{3} u\,\mathrm{arctan}{\,u}-\frac{1}{2}\ln{(u^2+1)} $$

• Substitute back \(u=3x,\) simplify, and add the constant of integration.

$$\begin{align} \int \mathrm{arctan}{\, 3x} \, \mathrm{d}x &= \frac{1}{3} \left(3x \,\mathrm{arctan}{\,(3x)}-\frac{1}{2}\ln{((3x)^2+1)}\right) \\ &= x\,\mathrm{arctan}{\, 3x} - \frac{1}{6}\ln{\left( 9x^2+1\right)} + C \end{align}$$