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Integration using Inverse Trigonometric Functions

- Calculus
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You might have heard of some Integration techniques, like Integration by Substitution, Integration by Parts, and Trigonometric Substitution. These techniques allow us to tackle integration, which is not a straightforward operation.

The Trigonometric Substitution method involves clever substitutions and trigonometric identities, but choosing the correct substitution is not always crystal clear. Luckily, we can take advantage of inverse trigonometric functions' derivatives to identify some functions' antiderivatives. Here we will discuss Integration using Inverse Trigonometric Functions.

In order to integrate using Inverse Trigonometric Functions, we need to know how to differentiate them. Let's take a quick refresher on this.

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!

What about the rest of the inverse trigonometric functions? Let's take a look at the inverse cosine function and its derivative.

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.

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

There are six inverse trigonometric functions, but we usually only use three of them as the other three are the same expressions, just with a negative sign. Let's take a look at which inverse trigonometric functions are used in integration.

The derivatives of the main inverse trigonometric functions used for integration are the following:

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

$$\frac{\mathrm{d}}{\mathrm{d}x}\arctan{\frac{x}{a}}=\frac{a}{a^2+x^2}$$

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

The above expressions are used to find the antiderivative of some rational functions.

The main antiderivatives that relate to inverse trigonometric functions are the following:

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

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

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

The signs matter! If the variable is being subtracted, you will likely have to use the inverse sine function. If the constant is being subtracted, then the inverse secant is more fitting. Finally, if the variable and the constant are being added, then it is time to use the inverse tangent.

Let's see an example using one of the above formulas.

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 \(a=3\).*

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

Pretty straightforward, right? Sometimes we will not have a perfect square, so we got to be clever.

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 \(a=\sqrt{5}\).*

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

Remember that these formulas can be combined along with other integration techniques. As usual, do not forget to add the constant of integration at the end!

We have seen integrals where inverse trigonometric functions are involved **in the answer.** But what if we want to integrate an inverse trigonometric function itself? We will now see how.

The antiderivatives of the inverse trigonometric functions are given as follows:

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

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

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

$$\int\mathrm{arccot}{\,x}\,\mathrm{d}x=x\,\mathrm{arccot}{\,x}+\frac{1}{2}\ln{(x^2+1)}+C$$

$$\int\mathrm{arcsec}{\,x}\,\mathrm{d}x=x\,\mathrm{arcsec}{\,x}-\ln{\big(x+\sqrt{x^2-1}\big)}+C$$

$$\int\mathrm{arccsc}{\,x}\,\mathrm{d}x=x\,\mathrm{arccsc}{\,x}+\ln{\big(x+\sqrt{x^2-1}\big)}+C$$

How do we prove the formulas for integrating the inverse trigonometric functions? If you take a closer look at the formulas, you will notice something common amongst them: every formula starts with \( x\) multiplying the respective inverse trigonometric function. How do we achieve this? Using Integration by Parts, of course! Let's see how this is done for the inverse sine function.

$$\int\arcsin{x}\,\mathrm{d}x$$

Here we will let \(u=\arcsin{x}\) and \(\mathrm{d}v=\mathrm{d}x\). We find \(\mathrm{d}u\) by differentiating the inverse sine function.

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

Remember that the integral of \(\mathrm{d}x\) is just \(x\).$$\mathrm{d}v=\mathrm{d}x \rightarrow v=x $$

We can now use the Integration by Parts formula \(\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u\).

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

The resulting integral can be evaluated using Integration by Substitution. Here we will let \(b=1-x^2\) (we use \(b\) to avoid confusion) so \(\mathrm{d}x=\text{-}\frac{1}{2}\mathrm{d}b\).

$$\int\arcsin{x}\,\mathrm{d}x=x\,\arcsin{x}+\frac{1}{2}\int\frac{\mathrm{d}b}{\sqrt{b}}$$

We evaluate the last integral using the power rule.

$$\int\arcsin{x}\,\mathrm{d}x=x\,\arcsin{x}+\sqrt{b}+C$$

And finally, we substitute back \(b=1-x^2\).

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

Similar steps are used when proving the rest of the antiderivatives of inverse trigonometric functions.

We will now take a look at some examples of integrals that involve inverse trigonometric functions in any way.

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 \(u=x^2\).*

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

Let's see an example involving the inverse secant function.

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.

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

It's time for an example of the integral of an inverse trigonometric function.

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.

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

- The main
**Integrals Resulting in Inverse Trigonometric Functions**are the following:- \(\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\arcsin{\frac{x}{a}}+C\)
- \(\int\frac{\mathrm{d}x}{a^2+x^2}=\frac{1}{a}\arctan{\frac{x}{a}}+C\)
- \(\int\frac{\mathrm{d}x}{x\sqrt{x^2-a^2}}=\frac{1}{a}\mathrm{arcsec}{\frac{x}{a}}+C\)

- The inverse trigonometric functions can be integrated using
**Integration by Parts**. - The antiderivatives of the inverse trigonometric functions are the following:
- \(\int\arcsin{x}\,\mathrm{d}x=x\arcsin{x}+\sqrt{1-x^2}+C\)
- \(\int\arccos{x}\,\mathrm{d}x=x\arccos{x}-\sqrt{1-x^2}+C\)
- \(\int\mathrm{arctan}{\,x}\,\mathrm{d}x=x\,\mathrm{arctan}{\,x}-\frac{1}{2}\ln{(x^2+1)}+C\)
- \(\int\mathrm{arccot}{\,x}\,\mathrm{d}x=x\,\mathrm{arccot}{\,x}+\frac{1}{2}\ln{(x^2+1)}+C\)
- \(\int\mathrm{arcsec}{\,x}\,\mathrm{d}x=x\,\mathrm{arcsec}{\,x}-\ln{\big(x+\sqrt{x^2-1}\big)}+C\)
- \(\int\mathrm{arccsc}{\,x}\,\mathrm{d}x=x\,\mathrm{arccsc}{\,x}+\ln{\big(x+\sqrt{x^2-1}\big)}+C\)

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