5.7 Further Transcendental Functions

Recall that

\[\boxed{\bbox[#FAF8ED,5pt]{\log_e x = \ln x := \int \limits_1^x \dfrac{dt}{t} \hskip 20pt \mbox{for} \ x>0}}\tag {1}\]

Thus, by FTC I,

\[ \int \limits_a^b \dfrac{dx}{x} = \ln \dfrac{b}{a} \]

Figure 5.48: Use the slider to see how the area under the graph changes.

In a similar fashion, we can express \(\sin^{-1} x\) as a definite integral using the derivative formula from Section 3.9 (Figure 5.49):

Figure 5.49: Use the slider to see how the area under the graph changes. In Section 7.6 we will see that this even works for \(x=1\) even though \(\lim \limits_{x \rightarrow 1} \frac{1}{\sqrt{1-t^2}}=\infty\).

Since sin−1 0 = 0, we have

On the other hand, the derivative formulas from Section 3.8 yield integration formulas that are useful for evaluating new types of integrals.

Inverse Trigonometric Functions

It is possible (and mathematically, it is more efficient) to take Eq. (1) as the definition of ln x and to define ex as the corresponding inverse function (see Exercises 78–79).

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In this list, we omit the integral formulas corresponding to the derivatives of cos−1x, cot−1x, and csc−1x because the integrals differ only by a minus sign from those already on the list. For example,

EXAMPLE 1

Evaluate .

Solution This integral is the area of the region in Figure 5.50. By Eq. (3),

Figure 5.50: The shaded region has an area equal to .

EXAMPLE 2 Using Substitution

Evaluate

Solution Notice that can be written as , so it makes sense to try the substitution u = 2x, du = 2 dx. Then

The new limits of integration are . By Eq. (4),

In substitution, we usually define u as a function of x. Sometimes, it is more convenient to define x as a function of u. We do this here, where we set x = 2u.

EXAMPLE 3 Using Substitution

Evaluate .

Solution Let us first rewrite the integrand:

Thus it makes sense to use the substitution . Then and

The new limits of integration are u (0) = 0 and :

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Integrals Involving \(f(x)=b^x\)

The exponential function f(x) = ex is particularly convenient because ex is both its own derivative and its own antiderivative. For other bases \(b>0\), we have

This translates into the integral formula

REMINDER \[b=e^{\ln b}, b^x=e^{(\ln b) x}\]

EXAMPLE 4

Evaluate .

Solution Apply Eq. (5) with b = 7.

EXAMPLE 5

Evaluate .

Solution Use the substitution u = sin θ, du = cos θ dθ. The new limits of integration become u (0) = 0 and u(π/2) = 1:

5.7.1 Summary

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