# Inverse Trigonometric Functions

In this section we will introduce the inverse
trigonometric functions . We will begin with

the inverse tangent function since, as indicated in Section 6.4, we need it to
complete the

story of the integration of rational functions .

Strictly speaking, the tangent function does not have an inverse. Recall that in
order

for a function f to have an inverse function, for every y in the range of f
there must be

exactly one x in the domain of f such that f(x) = y. This is false for the
tangent function

since, for example, both tan(0) = 0 and tan(π ) = 0. In fact, since the
tangent function is

periodic with period π, if tan(x) = y, then tan(x + nπ ) = y for any
integer n. However,

the tangent function is increasing on the interval
, taking on every value in its

range (−∞,∞) exactly once. Hence we may define an inverse for the tangent
function

if we consider it with the restricted domain .
That is, we will define an inverse

tangent function so that it takes on only values in
.

**Definition **The arc tangent function, with value at x denoted by either
arctan(x) or

tan^{-1}(x), is the inverse of the tangent function with restricted domain
.

In other words, for ,

y = tan^{-1}(x) if and only if tan(y) = x.
(6.5.1)

For example, , and
. In particular, note that

even though since 0 is between
and , but
π is not between

and .

The domain of the arc tangent function is (−∞,∞), the range of the tangent
function,

and the range of the arc tangent function is ,
the domain of the restricted tangent

function. Moreover, since

and

we have

and

Figure 6.5.1 Graph of y = tan^{-1}(x)

Hence and
are horizontal asymptotes for the graph of y = tan^{-1}(x), as

shown in Figure 6.5.1.

To differentiate the arc tangent function we imitate the method we used to
differentiate

the logarithm function . Namely, if y = tan^{-1}(x), then tan(y) = x, so

Hence

from which it follows that

Now

so we have

Hence we have demonstrated the following proposition.

**Proposition**

As a consequence of the proposition, we also have

Note that 1+x^{2} is an irreducible quadratic polynomial. We
will see more examples of

this type in the following examples.

**Example** Using the chain rule, we have

**Example** Evaluating is
similar to evaluating . That is, we will

use integration by parts with

Then

Using the substitution

we have , from which it follows that

Thus

**Example** To evaluate ,
we make the substitution

Then , so

**Example** To evaluate ,
we first note that x^{2} + x + 1 does not factor,

that is, is irreducible, and so we cannot use a partial fraction decomposition .
In general,

a quadratic polynomial ax ^{2} + bx + c is irreducible if b^{2} −
4ac < 0 since, in that case, the

quadratic formula yields complex solutions for the equation ax^{2}+bx+c = 0. For
x^{2}+x+1

we have b^{2} − 4ac = −3. In this case it is helpful to simplify the function
algebraically by

completing the square of the denominator , thus making the problem similar to the
previous

example. That is, since

we have

Now we can make the substitution

Then , so

**Partial fraction decomposition: Irreducible quadratic
factors
**The last two examples illustrate techniques that we may use to evaluate the
integral of

a rational function with an irreducible quadratic polynomial in the denominator. With

this we are now in a position to consider the final case of partial fraction decomposition.

Specifically, suppose we want to evaluate

where f and g are both polynomials and the degree of f is
less than the degree of g.

Moreover, suppose that (ax^{2} + bx + c)^{n} is a factor of g, where n is a positive
integer and

ax^{2} + bx + c is irreducible. Then the partial fraction
decomposition of must contain

a sum of terms of the form

where and
are constants. Note that the terms in the
partial

fraction decomposition corresponding to an irreducible quadratic factor differ
from the

terms for a linear factor in that the numerators of the terms in (6.5.6) need
not be constants,

but may be first degree polynomials themselves. As before, this is best
illustrated with an

example.

**Example** To evaluate we need to find
constants A, B, and C such that

Combining the terms on the right, we have

Hence

Equating the coefficients of the polynomials on the left
and right gives us the system of

equations

Thus B = −1 and

Hence

where the final integral follows from the substitution u =
1 + x^{2} as in an earlier example.

If, unlike this example , the partial fraction decomposition of
results in a term of

the form

where n > 1 and ax^{2} + bx + c is irreducible,
then the integration may still be difficult to

carry out, perhaps even requiring some of the ideas of trigonometric
substitutions that we

will discuss in the next section. However, there is a limit to what should be
done without

the aid of a computer, or at least a table of integrals. There is a point after
which some

integrations become so complicated and time-consuming that in practice they
should be

given to a computer algebra system .

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