# Calculus With Inverse Trigonometric Functions

How would you like it if you could evaluate \( \displaystyle \int \dfrac{dx}{\sqrt{1-x^{2}}} \) in your head, without doing trigonometric substitutions? How about \( \displaystyle \int \dfrac{dx}{1+x^2 } \)? This wiki will show you how. It turns out that as you would use a trigonometric or hyperbolic trigonometric substitution to evaluate those integrals, the results of those integrals invariably result in inverse trigonometric or inverse hyperbolic functions.

#### Contents

## Derivatives of Inverse Trigonometric Functions

Remember what the inverse of a function is? Let's define the inverses of trigonometric functions such as \( y = \sin x \) by writing \( x = \sin y\), which is the same as \( y= \sin^{-1} x \) or \( y = \arcsin x \). You can apply this convention to get other inverse trig functions. So, if we want to differentiate inverse functions, we can use the formula \[ \left(f^{-1}\right)'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)}. \] Another fact that will be used is \( \sin^{2} x + \cos^{2} x = 1\), which is an application of the Pythagorean theorem to the unit circle.

\[ \dfrac{d}{dx} (\arcsin x) =? \]

Using the formula of derivatives for inverse functions, \( \left(f^{-1}\right)'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)} \), if we say \( y = \arcsin(x)\), we can write \[ \dfrac{d}{dx} \left[ \arcsin(x) \right] = \dfrac{1}{\dfrac{d}{dy}\left[ \sin y \right]} \] because the inverse of \( y = \sin x \) is \(x = \sin y\), and then we can rewrite as \(\dfrac{1}{ \cos y } \). But what was \(y \)? \( y = \arcsin x\), so we can then say \[ \dfrac{d}{dx}( \arcsin x ) = \dfrac{1}{\cos\left(\arcsin x \right) } = \dfrac{1}{\sqrt{1-\sin^{2} \left( \arcsin x \right)}} = \dfrac{1}{\sqrt{1-x^{2}}}.\ _\square \]

Now that I have found the derivative of one of the inverse trigonometric functions, you should find the other five by yourself. The answers are posted below:

\[\begin{align} \dfrac{d}{dx} \left[\sin^{-1} x \right] &= \dfrac{1}{\sqrt{1-x^2}} \\ \dfrac{d}{dx} \left[ \cos^{-1} x \right] &= \dfrac{-1}{\sqrt{1-x^{2}}} \\ \dfrac{d}{dx} \left[ \tan^{-1} x \right] &= \dfrac{1}{1+x^2 }\\ \dfrac{d}{dx} \left[ \cot^{-1} x \right] &= \dfrac{-1}{1+x^{2}} \\ \dfrac{d}{dx} \left[ \sec^{-1} x \right] &= \dfrac{1}{\left|x \right| \sqrt{x^{2} - 1 }} \\ \dfrac{d}{dx} \left[ \csc^{-1} x \right] &= \dfrac{-1}{\left| x\right| \sqrt{x^{2} -1}}. \end{align} \]

**Questions to consider:**

- Why do we need the absolute value signs around \(x \) for the derivatives of inverse secant and cosecant?

## Derivatives of Inverse Hyperbolic Trigonometric Functions

Remember that we can say that \( \displaystyle \int \dfrac{dx}{\sqrt{1-x^{2}} } = \arcsin x + C \)? What enabled us to say that? We used the formula of derivatives for inverse functions, \( \left(f^{-1}\right)'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)} \), and the Pythagorean identities.

But what if we wanted to find \( \displaystyle \int \dfrac{dx}{\sqrt{1+x^{2}}} \)? Our "normal" inverse trigonometric functions don't provide a way for us to evaluate this. However, we could notice the fact that hyperbolic trigonometric function identities are very similar to normal trigonometric identities with minor sign changes. Why don't we take a look at the Pythagorean identity for hyperbolic trigonometric functions.

We know that \( \sinh x = \dfrac{e^{x} - e^{-x}}{2} \) and \(\cosh x = \dfrac{e^{x} + e^{-x}}{2} \).

What if we evaluate \( \left(\dfrac{e^{x} + e^{-x}}{2} \right)^{2} - \left(\dfrac{e^{x} - e^{-x}}{2} \right)^{2} \)?

Then the answer is \( \left( \dfrac{e^{2x} + e^{-2x} + 2}{4 }\right) - \left( \dfrac{e^{2x} + e^{-2x} - 2}{4} \right) = 1 .\)

Because \( \sinh x = \dfrac{e^{x} - e^{-x}}{2} \) and \(\cosh x = \dfrac{e^{x} + e^{-x}}{2} \), we have just proven that

\[\cosh^{2} x - \sinh^{2} x = 1.\]

Using this fact, let's now use the formula for derivatives of inverse functions and the Pythagorean identity for hyperbolic trigonometric functions to derive the formulas for derivatives of hyperbolic trigonometric functions.

Let's find the derivative of the following functions: \[\begin{array} &y= \sinh^{-1} x, &y= \cosh^{-1} x, &y = \tanh^{-1} x, &y= \text{coth}^{-1}, &y= \text{sech}^{-1}, &y=\text{csch}^{-1} .\end{array}\]

\[ \dfrac{d}{dx} \left[ \sinh^{-1} x \right] = ? \]

Using \( \left(f^{-1}\right)'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)} \), if we say that \( y = \sinh^{-1} x \), we can write

\[y' = \dfrac{1}{\dfrac{d}{dy}( \sinh y ) } = \dfrac{1}{\cosh y }.\]

But what was \(y\)? \[ y' = \dfrac{1}{\cosh\left(\sinh^{-1} x \right) }. \]

Knowing that \( \cosh^{2} x - \sinh^{2} x = 1 \), we can say that \( \cosh x = \sqrt{1+ \sinh^{2} x} \). So then, we can write

\[ y' = \dfrac{1}{\sqrt{1+ \sinh^{2} \left( \sinh^{-1} x \right) } } = \dfrac{1}{\sqrt{1+x^{2}}}.\ _\square \]

Now that I have done one example for you, please derive the rest of the formulas (answers will be shown below) for the derivatives of the rest of the inverse hyperbolic trigonometric functions:

\[\begin{align} \dfrac{d}{dx} \left[ \sinh^{-1} x\right] &= \dfrac{1}{\sqrt{1+x^{2}}} \\ \dfrac{d}{dx} \left[\cosh^{-1} x \right] &= \dfrac{1}{\sqrt{x^{2} - 1}} \\ \dfrac{d}{dx} \left[ \tanh^{-1} x \right] &= \dfrac{1}{1- x^{2}}\\ \dfrac{d}{dx} \left[\text{coth}^{-1} x \right] &= \dfrac{1}{1- x^{2}}\\ \dfrac{d}{dx} \left[ \text{sech}^{-1} x \right] &=-\dfrac{1}{x\sqrt{1-x^{2}}}\\ \dfrac{d}{dx} \left[\text{csch}^{-1} x \right] &= -\dfrac{1}{\left|x\right| \sqrt{1+x^{2}}}. \end{align}\]

**Questions to consider:**

- Why are the derivatives of \( \tanh^{-1} x \) and \( \text{coth}^{-1} x \) seemingly the same? When would you know to say \( \tanh^{-1} x \) vs. \( \text{coth}^{-1} x \)?
- For the other derivative formulas, are you allowed to use them for all values of \(x\)? Why or why not?
- Why does the \(x \) outside the radical in the derivative of hyperbolic cosecant inverse have the absolute value sign, while the \(x \) outside the radical in the derivative of hyperbolic secant inverse does not?

Evaluate \( \dfrac{d}{dx} \left[ \text{sech}^{-1} \left( x^{2016} \right) \right] \).

## Integrals Resulting in Inverse Trigonometric Functions

Now, let's do the *inverse* of what we just did (please laugh). We know that integration is the opposite of differentiation. So, we can immediately write the following:

\[\begin{align} \displaystyle \int \dfrac{dx}{\sqrt{1-x^{2}}} &= \sin^{-1} x + C \\ \int - \dfrac{dx}{\sqrt{1-x^{2}}} &= \cos^{-1} x + C \\ \int \dfrac{dx}{1+x^{2} } &= \tan^{-1} x + C \\ \int - \dfrac{dx}{1+x^{2}} &= \cot^{-1} x + C\\ \int \dfrac{dx}{x \sqrt{x^{2} - 1}} &= \sec^{-1} \left|x \right| + C \\ \int -\dfrac{dx}{x \sqrt{x^{2} - 1}} &= \csc^{-1} \left| x \right| +C. \end{align}\]

**Questions to consider:**

- Isn't \( \displaystyle \int - \dfrac{dx}{\sqrt{1-x^{2}}} \) the same as \( \displaystyle - \int \dfrac{dx}{\sqrt{1-x^{2}}} \), which would just result in \( - \sin^{-1} x + C \)?
- Why do we move the \( \left| x \right| \) from outside the radical sign to the inside of secant inverse and cosecant inverse \( \left( \sec^{-1} \left| x \right| \right. \) and \( \left. \csc^{-1} \left| x \right| \right) \)? Is that even allowed?

Evaluate \( \displaystyle \int \dfrac{dx}{\sqrt{9-4x^{2}}} \).

Solution:We want to get our integrand in the form \( \dfrac{du}{\sqrt{1-u^{2}}} \) because that's something we can integrate without doing trigonometric substitutions.

Factor out the 9, then we can write \[ \displaystyle \int \dfrac{dx}{\sqrt{9\left(1- \dfrac{4}{9}x^{2} \right)}} = \int \dfrac{dx}{3\sqrt{1- \dfrac{4}{9}x^{2} }} . \qquad (1)\] Now, set \( u = \dfrac{2}{3} x\) and \(\dfrac{du}{dx} = \dfrac{2}{3} \). Why do we do this? Because we want an integrand in the form \( \dfrac{du}{\sqrt{1-u^{2}}} \). Then \((1)\) becomes \[\dfrac{3}{2} \cdot \dfrac{1}{3} \displaystyle \int \dfrac{du}{\sqrt{1-u^{2}}} .\] which looks really familiar and is equivalent to \[ \dfrac{1}{2}\arcsin u + C, \] where \(C\) is the constant of integration.

But we're not done. What was \(u \)? \( u = \dfrac{2}{3} x \). We have to plug that back in (unless we were doing a definite integral, and had found new upper and lower limits of integration for the u-substitution). Then the final result comes out to \[ \dfrac{1}{2} \arcsin \left( \dfrac{2}{3}x \right) + C.\ _\square \]

## Integrals Resulting in Inverse Hyperbolic Trigonometric Functions

Just as we did for the inverse trigonometric functions, we can write some integrals right off the bat:

\[\begin{align} \int \dfrac{dx}{\sqrt{1+x^{2}}} &= \sinh^{-1} x + C \\ \int \dfrac{dx}{\sqrt{x^{2} - 1}} &= \cosh^{-1} x + C \\ \int \dfrac{dx}{1-x^{2} } &= \tanh^{-1} x + C \qquad \left(\text{or }~ \text{coth}^{-1} x + C\right) \\ \int \dfrac{dx}{x \sqrt{1-x^{2}}} &= - \text{sech}^{-1} x + C \\ \int \dfrac{dx}{x \sqrt{1+x^{2}}} &= - \text{csch}^{-1} x + C. \end{align}\]

But that's actually not the whole story. How can we rewrite \(y= \sinh^{-1} x \)? We can say \( \dfrac{ e^{y}-e^{-y}}{2} = x \). What if we solve for \( y\)? Then we have

\[\begin{align} e^{y} - e^{-y} &= 2x \\ e^{y} \left( e^{y} - e^{-y} \right) &= 2x e^{y} \\ e^{2y} - 1 &= 2xe^{y}. \end{align} \]

Let \( u = e^{y} \), then

\[\begin{align} u^{2} - 2xu - 1 &= 0 \\ u &= \dfrac{2x \pm \sqrt{4x^{2} - 4(-1)}}{2} \\ &= x \pm \sqrt{x^{2} + 1}. \end{align} \]

Since we said \( u= e^{y} \), we can also say \( \ln u = y \) because we were looking for \(y\). But what was \(u \)? We solved a quadratic equation and said that \( u = x \pm \sqrt{x^{2} + 1} \). So can we say that \( \sinh^{-1} x = \ln \left( x \pm \sqrt{x^{2} + 1} \right) \)? No! What condition must we impose on the argument of a logarithm? It must be positive! If we look at the graph of \( y = x + \sqrt{x^{2} + 1}, \) we're okay because it's always positive. However, if we look at the graph of \( y = x - \sqrt{x^{2} + 1} \), we're not okay because it's always negative. So, if we can say that \( \sinh^{-1} x = \ln \left( x + \sqrt{x^{2} + 1} \right) \), then we can also say that \( \displaystyle \int \dfrac{dx}{\sqrt{x^{2} + 1 }} \) could also be written as \( \ln \left( x + \sqrt{x^{2} + 1} \right) + C \), where \(C\) is the constant of integration.

So, please rewrite all of the results of those integrals using natural logarithm. The answers are listed below:

\[\begin{align} \int \dfrac{dx}{\sqrt{1+x^{2}}} &= \ln \left( x + \sqrt{x^{2} + 1 } \right) + C \\ \int \dfrac{dx}{\sqrt{x^{2} - 1}} &= \ln \left(x + \sqrt{x^{2} -1 } \right) + C \\ \int \dfrac{dx}{1-x^{2} } &= \dfrac{1}{2} \ln \left| \dfrac{1+x}{1-x}\right| + C \\ \int \dfrac{dx}{x \sqrt{1-x^{2}}} &= - \ln \left( \dfrac{x+ \sqrt{1-x^{2}}}{\left| x \right|}\right)+ C \\ \int \dfrac{dx}{x \sqrt{1+x^{2}}} &= - \ln \left( \dfrac{1+\sqrt{1+ x^{2} }}{\left| x \right|} \right)+C. \end{align}\]

**Questions to consider:**

Regarding both sets of formulas, consider the following questions:

- Do all of those formulas work for all values of \(u\)? Why or why not? If necessary, state the restrictions of \(u\).
- Why are there absolute value signs around \( x \) in the results of the last two integrals?
- When do you use \( \tanh^{-1} x \) in the result of an integral? \( \text{coth}^{-1} x \)?

Evaluate \( \displaystyle \int \dfrac{dx}{\sqrt{4x^{2} - 9}} \).

We want to make our integral look something like \( \displaystyle \int \dfrac{du}{\sqrt{u^{2} - 1} } \).

Let's first factor out the \(9\):

\[\int \frac{dx}{\sqrt{9 \left( \frac{4}{9}x^{2} - 1 \right)}} = \frac{1}{3}\int \frac{dx}{\sqrt{ \frac{4}{9}x^{2} - 1 }}.\qquad (1) \]

Now, let's do \( u\)-substitution: let \( u = \dfrac{2}{3}x\) and \( du = \dfrac{2}{3} dx \). Then \((1)\) becomes

\[ \dfrac{1}{3} \cdot \dfrac{3}{2} \int \dfrac{du}{\sqrt{u^{2} - 1}},\]

which looks awfully familiar!

We have two choices: this last expression is equal to

\[ \dfrac{1}{2} \cosh^{-1} u + C ~~\text{ or }~~ \dfrac{1}{2} \ln \left(u + \sqrt{u^{2} - 1} + C\right), \]

where \(C\) is the constant of integration. But in either case, what was \(u\)? \( u = \dfrac{2}{3}x \).

Therefore, the given expression is equal to

\[ \dfrac{1}{2} \cosh^{-1} \left( \dfrac{2}{3} x \right) + C ~~\text{ or }~~ \dfrac{1}{2} \ln \left( \dfrac{2}{3}x + \sqrt{\dfrac{4}{9}x^{2} - 1} \right) + C. \ _\square \]

**Cite as:**Calculus With Inverse Trigonometric Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/applying-differentiation-rules-to-inverse/