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6.5: Derivatives and Integrals Involving Inverse Trigonometric Functions

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

A student will be able to:

  • Learn the basic properties inverse trigonometric functions.
  • Learn how to use the derivative formula to use them to find derivatives of inverse trigonometric functions.
  • Learn to solve certain integrals involving inverse trigonometric functions.

A Quick Algebraic Review of Inverse Trigonometric Functions

You already know what a trigonometric function is, but what is an inverse trigonometric function? If we ask what is \sin (\pi/6) equal to, the answer is (1/2). That is simple enough. But what if we ask what angle has a sine of (1/2)? That is an inverse trigonometric function. So we say \sin (\pi/6) = (1/2), but \sin^{-1} (1/2) = (\pi/6). The “\sin^{-1}” is the notation for the inverse of the sine function. For every one of the six trigonometric functions there is an associated inverse function. They are denoted by

\sin^{-1} x, \cos^{-1} x, \tan^{-1} x, \sec^{-1} x, \csc^{-1} x, \cot^{-1} x

Alternatively, you may see the following notations for the above inverses, respectively,

\text{arcsin} \ x, \text{arccos} \ x, \text{arctan} \ x, \text{arcsec} \ x, \text{arccsc} \ x, \text{arccot} \ x

Since all trigonometric functions are periodic functions, they do not pass the horizontal line test. Therefore they are not one-to-one functions. The table below provides a brief summary of their definitions and basic properties. We will restrict our study to the first four functions; the remaining two, \csc^{-1} and \cot^{-1}, are of lesser importance (in most applications) and will be left for the exercises.

Inverse Function Domain Range Basic Properties
 \sin^{-1}  -1 \le x \le 1  \frac{- \pi} {2} \le y \le \frac{\pi} {2}  \sin (\sin^{-1} (x)) = x
 \cos^{-1}  -1 \le x \le 1  0 \le y \le \pi  \cos^{-1} (\cos x) = \cos (\cos^{-1}(x)) = x
 \tan^{-1} all  R  \left (\frac{- \pi} {2}, \frac{\pi} {2}\right )  \tan^{-1} (\tan x) = \tan (\tan^{-1}(x)) = x
 \sec^{-1}  (-\infty, -1] \cup [1, +\infty)  \left [0, \frac{\pi} {2}\right ) \cup \left (\frac{\pi} {2} , \pi \right]  \sec^{-1} (\sec x) = \sec (\sec^{-1}(x)) = x

The range is based on limiting the domain of the original function so that it is a one-to-one function.

Example 1:

What is the exact value of  \sin^{-1} (\sqrt{3}/2)?


This is equivalent to  \sin x = \frac{\sqrt{3}} {2} . Thus  \sin^{-1} (\sqrt{3}/2) = \pi / 3. You can easily confirm this result by using your scientific calculator.

Example 2:

Most calculators do not provide a way to calculate the inverse of the secant function,  \sec^{-1} x. A practical trick however is to use the identity

\sec^{-1} x = \cos^{-1} \frac{1} {x}

(Recall that  \sec \theta = \frac{1} {\cos \theta}.)

For practice, use your calculator to find \sec^{-1} (3.24).



\frac{1} {x} = \frac{1} {3.24} = 0.3086,

\sec^{-1} 3.24 = \cos^{-1} 0.3086 = 72^\circ.

Here are two other identities that you may need to enter into your calculator:

\csc^{-1} x & = \sin^{-1} \frac{1} {x},\\\cot^{-1} x  & = \tan^{-1} \frac{1} {x}.

The Derivative Formulas of the Inverse Trigonometric Functions

If u is a differentiable function of x then the generalized derivative formulas for the inverse trigonometric functions are (we introduce them here without a proof):

\frac {d}{dx} \left [ {\sin^{-1}{u}} \right ] & = \frac {1}{\sqrt {1-u^2}} \frac {du}{dx}   {-1 < u < 1} \\\frac {d}{dx} \left [ {\cos^{-1}{u}} \right ] & = \frac {-1}{\sqrt {1-u^2}} \frac {du}{dx}   {-1 < u < 1} \\\frac {d}{dx} \left [ {\tan^{-1}{u}} \right ] & = \frac {1}{1+u^2} \frac {du}{dx}  {- \infty < x < \infty}   \\\frac {d}{dx} \left [ {\sec^{-1}{u}} \right ] & = \frac {1}{\begin{vmatrix}{u} \end{vmatrix} \sqrt {u^2-1}}  \frac {du}{dx} {\begin{vmatrix}{u} \end{vmatrix}} > 1  \\\frac {d}{dx} \left [ {\csc^{-1}{u}} \right ] & = \frac {-1}{\begin{vmatrix}{u} \end{vmatrix} \sqrt {u^2-1}} \frac {du}{dx} {\begin{vmatrix}{u} \end{vmatrix}} > 1  \\\frac {d}{dx} \left [ {\cot^{-1}{u}} \right ] & = \frac {-1}{1+u^2} \frac {du}{dx}   {- \infty < x < \infty}

Example 3:

Differentiate y = \sin^{-1} (2x^4)


Let u = 2x^4, so

\frac {dy}{dx} & = \frac {1}{\sqrt {1-(2x^4)^2}} \cdot (8x^3)\\& = \frac {8x^3}{\sqrt {1-4x^8}}.

Example 4:

Differentiate  \tan^{-1}(e^{3x}).


Let u = e^{3x}, so

\frac {dy}{dx} & = \frac {1}{{1+(e^{3x})^2}} \cdot {3e^{3x}}\\& = \frac {3e^{3x}}{1+e^{6x}}.

Example 5:

Find dy/dx if y = \cos^{-1} (\sin x).


Let u = \sin x.

\frac {dy}{dx}  & = \frac {-1}{\sqrt {1 - \sin^2 x}}\\& = \frac {-1} {\cos x}.

The Integration Formulas of the Inverse Trigonometric Functions

The derivative formulas in the box above yield the following integrations formulas for inverse trigonometric functions:

\int \frac {du}{\sqrt {1-u^2}} & = \sin^{-1} u + c\\\int \frac {du}{1+u^2} & = \tan^{-1} u + c\\\int \frac {du} {u \sqrt {u^2-1}} & = \sec^{-1} {\begin{vmatrix}{u} \end{vmatrix}}+ c

Example 6:

Evaluate  \int \frac {dx}{1+4x^2}.


Before we integrate, we use u-substitution. Let u = 2x (the square root of 4x^2). Then du = 2dx. Substituting,

\int \frac {dx}{1+4x^2} & =  \int \frac {1/2}{1+u^2} du \\& =  \frac {1}{2} \int \frac {1}{1+u^2} du \\& =  \frac {1}{2} \tan^{-1}u + c \\& =  \frac {1}{2} \tan^{-1}(2x) + c.

Example 7:

Evaluate  \int \frac {e^x}{\sqrt {1-e^{2x}}} dx.


We use u-substitution. Let u = e^x, so du = e^x dx. Substituting,

\int \frac {e^x}{\sqrt {1-e^{2x}}} dx  & =  \int \frac {e^x}{\sqrt {1-u^2}} \frac {du}{e^x} \\& =  \int \frac {1}{\sqrt {1-u^2}} du \\& =  \sin^{-1}u + c \\& =  \sin^{-1}(e^x) + c.

Example 8:

Evaluate the definite integral  \int_{{ \ln}2}^{{ \ln}({2}/ \sqrt {3})} \frac {e^{-x}}{\sqrt {1-e^{-2x}}} dx .


Substituting u = e^{-x}, du = -e^{-x} dx.

To change the limits,

x & = \ln 2 \rightarrow u = e^{-x} = e^{-\ln 2} = e^{\ln {1/2}} = \frac {1}{2},\\x & = { \ln}\left (\frac {2}{\sqrt {3}}\right )  \rightarrow u = e^{-x} = e^{-\ln 2/ \sqrt {3}} =  \frac {\sqrt {3}}{2}.

Thus our integral becomes

\int_{{ \ln}2}^{{ \ln}({2}/ \sqrt {3})} \frac {e^{-x}}{\sqrt {1-e^{-2x}}} dx  & =  \int_{ {1/2}}^{\sqrt{3}/2} \frac {u}{\sqrt {1- u^2}} \frac {-du}{u} \\& = -  \int_{ {1/2}}^{\sqrt{3}/2} \frac {1}{\sqrt {1- u^2}} du \\& = -  \Bigg [ {\sin^{-1} u} \Bigg ]_{1/2}^{\sqrt{3}/2} \\& = -  \left [ \sin^{-1} \left (\frac {\sqrt {3}}{2}\right ) - \sin^{-1} \left (\frac{1}{2}\right ) \right ]\\& = -  \left [ \frac {\pi}{3}-\frac {\pi}{6}  \right ]\\& = - \frac {\pi}{6}.

Multimedia Links

For a video presentation of the derivatives of inverse trigonometric functions (4.4), see Math Video Tutorials by James Sousa, The Derivatives of Inverse Trigonometric Functions (8:55).

For three presentations of integration involving inverse trigonometric functions (18.0), see Math Video Tutorials by James Sousa, Integration Involving Inverse Trigonometric Functions, Part 1 (7:39)

; Math Video Tutorials by James Sousa, Integration Involving Inverse Trigonometric Functions, Part 2 (6:39)

; This last video includes an example showing completing the square (19.0), Math Video Tutorials by James Sousa, Integration Involving Inverse Trigonometric Functions, Part 3 (6:18).

Review Questions

  1. Find dy/dx of y = \sec^{-1} x^2.
  2. Find  dy/dx of  y = \frac {1} {\tan^{-1} x}.
  3. Find dy/dx of y = \ln(\cos^{-1} x).
  4. Find dy/dx of y = \sin^{-1} e^{-4x}.
  5. Find dy/dx of y = \sin^{-1} (x^2 \ln x).
  6. Evaluate   \int \frac {dx}{\sqrt{9-x^2}}.
  7. Evaluate   \int_{1}^{3}\frac {dx}{\sqrt {x}(x+1)}.
  8. Evaluate   \int \frac {x-3}{x^2+1}dx.
  9. Evaluate   \int_{-\sqrt {3}}^{0} \frac {x}{1+x^2}dx.
  10. Given the points A(2,1) and B(5,4), find a point Q in the interval [2,5] on the x-axis that maximizes angle  \measuredangle AQB.

Review Answers

  1.  \frac {2}{x \sqrt {x^4-1}}
  2.  \frac {-1}{(1+x^2)(\tan^{-1} x)^2}
  3.  -\frac{1}{(\cos^{-1} x) {\sqrt {1-x^2}}}
  4.  \frac {-4e^{-4x}}{\sqrt {1-e^{-8x}}}
  5.  \frac {x+2x \ln x} {\sqrt {1-x^4(\ln x)^2}}
  6.  \sin^{-1} \left ( \frac{x}{3} \right )+ C
  7. \pi/6
  8.  \frac {1}{2} \ln(x^2 + 1) - 3 \tan^{-1} x + C
  9. -\ln 2
  10.  1 + 2 \sqrt {2}

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Date Created:

Feb 23, 2012

Last Modified:

Aug 19, 2015
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