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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(π/6)\begin{align*}\sin (\pi/6)\end{align*} equal to, the answer is (1/2).\begin{align*}(1/2).\end{align*} That is simple enough. But what if we ask what angle has a sine of (1/2)\begin{align*}(1/2)\end{align*}? That is an inverse trigonometric function. So we say sin(π/6)=(1/2),\begin{align*}\sin (\pi/6) = (1/2),\end{align*} but sin1(1/2)=(π/6).\begin{align*}\sin^{-1} (1/2) = (\pi/6).\end{align*} The “sin1\begin{align*}\sin^{-1}\end{align*}” 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

sin1x,cos1x,tan1x,sec1x,csc1x,cot1x\begin{align*}\sin^{-1} x, \cos^{-1} x, \tan^{-1} x, \sec^{-1} x, \csc^{-1} x, \cot^{-1} x\end{align*}

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

arcsin x,arccos x,arctan x,arcsec x,arccsc x,arccot x\begin{align*}\text{arcsin} \ x, \text{arccos} \ x, \text{arctan} \ x, \text{arcsec} \ x, \text{arccsc} \ x, \text{arccot} \ x\end{align*}

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\begin{align*}\csc^{-1}\end{align*} and cot1,\begin{align*}\cot^{-1},\end{align*} are of lesser importance (in most applications) and will be left for the exercises.

Inverse Function Domain Range Basic Properties
sin1\begin{align*} \sin^{-1}\end{align*} 1x1\begin{align*} -1 \le x \le 1\end{align*} π2yπ2\begin{align*} \frac{- \pi} {2} \le y \le \frac{\pi} {2}\end{align*} sin(sin1(x))=x\begin{align*} \sin (\sin^{-1} (x)) = x\end{align*}
cos1\begin{align*} \cos^{-1}\end{align*} 1x1\begin{align*} -1 \le x \le 1\end{align*} 0yπ\begin{align*} 0 \le y \le \pi\end{align*} cos1(cosx)=cos(cos1(x))=x\begin{align*} \cos^{-1} (\cos x) = \cos (\cos^{-1}(x)) = x\end{align*}
tan1\begin{align*} \tan^{-1}\end{align*} all R\begin{align*} R\end{align*} (π2,π2)\begin{align*} \left (\frac{- \pi} {2}, \frac{\pi} {2}\right )\end{align*} tan1(tanx)=tan(tan1(x))=x\begin{align*} \tan^{-1} (\tan x) = \tan (\tan^{-1}(x)) = x\end{align*}
sec1\begin{align*} \sec^{-1}\end{align*} (,1][1,+)\begin{align*} (-\infty, -1] \cup [1, +\infty)\end{align*} [0,π2)(π2,π]\begin{align*} \left [0, \frac{\pi} {2}\right ) \cup \left (\frac{\pi} {2} , \pi \right]\end{align*} sec1(secx)=sec(sec1(x))=x\begin{align*} \sec^{-1} (\sec x) = \sec (\sec^{-1}(x)) = x\end{align*}

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)\begin{align*} \sin^{-1} (\sqrt{3}/2)\end{align*}?

Solution:

This is equivalent to sinx=32\begin{align*} \sin x = \frac{\sqrt{3}} {2}\end{align*} . Thus sin1(3/2)=π/3\begin{align*} \sin^{-1} (\sqrt{3}/2) = \pi / 3\end{align*}. 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.\begin{align*} \sec^{-1} x.\end{align*} A practical trick however is to use the identity

sec1x=cos11x\begin{align*}\sec^{-1} x = \cos^{-1} \frac{1} {x}\end{align*}

(Recall that secθ=1cosθ.\begin{align*} \sec \theta = \frac{1} {\cos \theta}.\end{align*})

For practice, use your calculator to find sec1(3.24).\begin{align*}\sec^{-1} (3.24).\end{align*}

Solution:

Since

1x=13.24=0.3086,\begin{align*}\frac{1} {x} = \frac{1} {3.24} = 0.3086,\end{align*}

sec13.24=cos10.3086=72.\begin{align*}\sec^{-1} 3.24 = \cos^{-1} 0.3086 = 72^\circ.\end{align*}

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

csc1xcot1x=sin11x,=tan11x.\begin{align*}\csc^{-1} x & = \sin^{-1} \frac{1} {x},\\ \cot^{-1} x & = \tan^{-1} \frac{1} {x}.\end{align*}

## The Derivative Formulas of the Inverse Trigonometric Functions

If u\begin{align*}u\end{align*} is a differentiable function of x\begin{align*}x\end{align*} then the generalized derivative formulas for the inverse trigonometric functions are (we introduce them here without a proof):

ddx[sin1u]ddx[cos1u]ddx[tan1u]ddx[sec1u]ddx[csc1u]ddx[cot1u]=11u2dudx1<u<1=11u2dudx1<u<1=11+u2dudx<x<=1|u|u21dudx|u|>1=1|u|u21dudx|u|>1=11+u2dudx<x<\begin{align*}\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}\end{align*}

Example 3:

Differentiate y=sin1(2x4)\begin{align*}y = \sin^{-1} (2x^4)\end{align*}

Solution:

Let u=2x4,\begin{align*}u = 2x^4,\end{align*} so

dydx=11(2x4)2(8x3)=8x314x8.\begin{align*}\frac {dy}{dx} & = \frac {1}{\sqrt {1-(2x^4)^2}} \cdot (8x^3)\\ & = \frac {8x^3}{\sqrt {1-4x^8}}.\end{align*}

Example 4:

Differentiate tan1(e3x).\begin{align*} \tan^{-1}(e^{3x}).\end{align*}

Solution:

Let u=e3x,\begin{align*}u = e^{3x},\end{align*} so

dydx=11+(e3x)23e3x=3e3x1+e6x.\begin{align*}\frac {dy}{dx} & = \frac {1}{{1+(e^{3x})^2}} \cdot {3e^{3x}}\\ & = \frac {3e^{3x}}{1+e^{6x}}.\end{align*}

Example 5:

Find dy/dx\begin{align*}dy/dx\end{align*} if y=cos1(sinx).\begin{align*}y = \cos^{-1} (\sin x).\end{align*}

Solution:

Let u=sinx.\begin{align*}u = \sin x.\end{align*}

dydx=11sin2x=1cosx.\begin{align*}\frac {dy}{dx} & = \frac {-1}{\sqrt {1 - \sin^2 x}}\\ & = \frac {-1} {\cos x}.\end{align*}

## The Integration Formulas of the Inverse Trigonometric Functions

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

du1u2du1+u2duuu21=sin1u+c=tan1u+c=sec1|u|+c\begin{align*}\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 \end{align*}

Example 6:

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

Solution:

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*}

Solution:

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*}.

Solution:

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*}

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)

; 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*}.

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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