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6.5: Derivatives and Integrals Involving Inverse Trigonometric Functions

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 $\sin^{-1} (\sqrt{3}/2)$?

Solution:

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 $\sec^{-1} (3.24).$

Solution:

Since

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

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

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

Solution:

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

Solution:

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 $y = \cos^{-1} (\sin x).$

Solution:

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

Solution:

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

Solution:

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 $\int_{{ \ln}2}^{{ \ln}({2}/ \sqrt {3})} \frac {e^{-x}}{\sqrt {1-e^{-2x}}} dx$.

Solution:

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

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

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 $dy/dx$ 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 $\measuredangle AQB$.

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. $1 + 2 \sqrt {2}$

Feb 23, 2012

Apr 01, 2014