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6.5: Derivatives and Integrals Involving Inverse Trigonometric Functions

Difficulty Level: At Grade Created by: CK-12
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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(π/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(π/6)=(1/2), but sin1(1/2)=(π/6). The “sin1” 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


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

arcsin x,arccos x,arctan x,arcsec x,arccsc x,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, csc1 and cot1, are of lesser importance (in most applications) and will be left for the exercises.

Inverse Function Domain Range Basic Properties
sin1 1x1 π2yπ2 sin(sin1(x))=x
cos1 1x1 0yπ cos1(cosx)=cos(cos1(x))=x
tan1 all R (π2,π2) tan1(tanx)=tan(tan1(x))=x
sec1 (,1][1,+) [0,π2)(π2,π] sec1(secx)=sec(sec1(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 sin1(3/2)?


This is equivalent to sinx=32 . Thus sin1(3/2)=π/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, sec1x. A practical trick however is to use the identity


(Recall that secθ=1cosθ.)

For practice, use your calculator to find sec1(3.24).





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


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):


Example 3:

Differentiate y=sin1(2x4)


Let u=2x4, so


Example 4:

Differentiate tan1(e3x).


Let u=e3x, so


Example 5:

Find dy/dx if y=cos1(sinx).


Let u=sinx.


The Integration Formulas of the Inverse Trigonometric Functions

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


Example 6:

Evaluate \begin{align*} \int \frac {dx}{1+4x^2}.\end{align*}


Before we integrate, we use \begin{align*}u-\end{align*}substitution. Let \begin{align*}u = 2x\end{align*} (the square root of \begin{align*}4x^2\end{align*}). Then \begin{align*}du = 2dx.\end{align*} Substituting,

\begin{align*}\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.\end{align*}

Example 7:

Evaluate \begin{align*} \int \frac {e^x}{\sqrt {1-e^{2x}}} dx. \end{align*}


We use \begin{align*}u-\end{align*}substitution. Let \begin{align*}u = e^x,\end{align*} so \begin{align*}du = e^x dx.\end{align*} Substituting,

\begin{align*}\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.\end{align*}

Example 8:

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


Substituting \begin{align*}u = e^{-x},\end{align*} \begin{align*}du = -e^{-x} dx.\end{align*}

To change the limits,

\begin{align*}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}.\end{align*}

Thus our integral becomes

\begin{align*}\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}.\end{align*}

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

Review Answers

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

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Date Created:
Feb 23, 2012
Last Modified:
Jun 10, 2016
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