Calculus I 05.07 Inverse Trigonometric Functions: Integration

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5.7 Inverse Trigonometric Functions: Integration

  • Integrate functions whose antiderivatives involve inverse trigonometric functions.
  • Use Commpleting the Square to integrate a function.

Integrals Involving Inverse Trigonometric Functions

The derivatives for the six inverse trigonometric functions fall into three pairs. In each pair, the derivative for one function is the negative for the other. For example,

$$\frac{d}{dx} \left [ \arcsin x \right ] = \frac{1}{\sqrt{1-x^2}}$$

and

$$\frac{d}{dx} \left [ \arccos x \right ] = -\frac{1}{\sqrt{1-x^2}}.$$

When listing the antiderivative that corresponds to each inverse trigonometric functions, only one from each pair is needed. The conventional notation is using \(\arcsin x\) as the antiderivative for \(1/\sqrt{1-x^2}\), rather than \(-\arccos x\). Theorem 5.7.1 gives one antiderivative formula for each in the three pairs. The proofs are left as an exercise.

Theorem 5.7.1 Integrals Involving Inverse Trigonometric Functions

Let \(u\) be a differentiable function for \(x\), and let \(a>0\).

$$\textbf{1.}\:\: \int \frac{du}{\sqrt{a^2-u^2}}= \arcsin \frac{u}{a}+C$$
$$\textbf{2.}\:\: \int \frac{du}{\sqrt{a^2+u^2}}= \frac{1}{a}\arctan \frac{u}{a}+C$$
$$\textbf{3.}\:\: \int \frac{du}{u\sqrt{a^2-u^2}}= \frac{1}{a} \operatorname{arcsec} \frac{\left | u \right |}{a}+C$$
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Example 5.7.1 Integration with Inverse Trigonometric Functions

$$\textbf{a.}\: \int \frac{dx}{\sqrt{4-x^2}}$$

$$= \arcsin \frac{x}{2}+C$$

$$\textbf{b.}\: \int \frac{dx}{2+9x^2}$$

$$=\frac{1}{3} \int \frac{3dx}{(\sqrt{2})^2+(3x)^2}$$

    \(u=3x,\:a=\sqrt{2}\)

$$=\frac{1}{3\sqrt{2}} \arctan \frac{3x}{\sqrt{2}}+C$$

$$\textbf{c.}\: \int \frac{dx}{x\sqrt{4x^2-9}}$$

$$=\int \frac{2dx}{2x\sqrt{(2x)^2-3^2}}$$

    \(u=2x,\:a=3\)

$$=\frac{1}{3} \operatorname{arcsec} \frac{\left | 2x \right |}{3}+C$$

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Example 5.7.2 Integration by Substitution

Find

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

SolutionThis integral does not fit any inverse trigonometric formula. Using the substitution \(u=e^x\), however, produces

$$u=e^x \: \rightarrow \: du=e^x\:dx \rightarrow dx=\frac{du}{e^x}=\frac{du}{u}.$$

With this substitution, integration proceeds as below.

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

$$=\int \frac{dx}{\sqrt{(e^{x})^2-1}}$$

    Write \(e^{2x}\) as \((e^{x})^2\)

$$=\int \frac{du/u}{\sqrt{u^2-1}}$$

    Substitute

$$=\int \frac{du}{u\sqrt{u^2-1}}$$

    Rewrite to fit the Arcsecant Rule

$$= \operatorname{arcsec} \frac{\left | u \right |}{1}+C$$

    Apply Arcsecant Rule

\(=\operatorname{arcsec} e^x+C\)

    Back-substitute
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A symbolic integration program can integrate functions such as the one in Example 5.7.2. In some cases, however, the program may fail to find an antiderivative for two reasons. First, some elementary functions do not have antiderivatives that are elementary functions. Second, every program has limitations, and the one entered might not be one the program can handle. Also remember that antiderivatives involving trigonometric functions or logarithmic functions can be written in many different forms. For example, one program found the integral in Example 5.7.2 as

$$\int \frac{dx}{\sqrt{e^{2x}-1}} = \arctan \sqrt{e^{2x}-1}+C.$$

Proving that this antiderivative is equivalent to the one found in Example 5.7.2 is left as an exercise.

Example 5.7.3 Rewriting as the Sum of Two Quotients

Find

$$\int \frac{x+2}{\sqrt{4-x^2}}\:dx.$$

Solution This integral does not appear to fit any basic integration formula. By splitting the integrand into two parts, however, the first part can be found with the Power Rule and the second part yields an inverse sine function.

$$\int \frac{x+2}{\sqrt{4-x^2}}\:dx$$

$$=\int \int \frac{x}{\sqrt{4-x^2}}\:dx+ \int \frac{2}{\sqrt{4-x^2}}\:dx$$

$$=-\frac{1}{2} \int (4-x^2)^{-1/2}(-2x)\:dx + 2 \int \frac{1}{\sqrt{4-x^2}}\:dx$$

$$= - \frac{1}{2} \left [ \frac{ (4-x^2)^{1/2}}{1/2} \right ] +2 \arcsin \frac{x}{2} + C$$

$$=-\sqrt{4-x^2}+2 \arcsin \frac{x}{2}+C$$

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Complete the Square

Completing the Square is a formula for integrating quadratic functions. For example, the quadratic \(x^2+bx+c\) can be written as the difference between two squares by adding and subtracting \((b/2)^2\).

$$x^2+bx+c=x^2+bx+\left ( \frac{b}{2} \right )^2 -\left ( \frac{b}{2} \right )^2+c=\left (x+ \frac{b}{2} \right )^2 - \left ( \frac{b}{2} \right )^2 +c$$

Example 5.7.4 Completing the Square

Find

$$\int \frac{dx}{x^2-4x+7}.$$

Solution Write the denominator as a sum with two squares, as shown below.

\(x^2-4x+7=(x^2-4x+4)-4+7=(x-2)^2+3=u^2+a^2\)

To complete the square form, let \(u=x-2\) and \(a=\sqrt{3}\).

$$\int \frac{dx}{x^2-4x+7}=\int \frac{dx}{(x-2)^2+3}= \frac{1}{\sqrt{3}} \arctan \frac{x-2}{\sqrt{3}}+C$$
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When the leading coefficient is not 1, it helps to factor before completing the square. For example, you can complete the square for \(2x^2-8x+10\) by factoring first.

\(2x^2-8x+10\)

\(=2(x^2-4x+5)\)

\(=2(x^2-4x+4-4+5\)

\(=2 \left [ (x-2^2)+1 \right ]\)

To complete the square when the coefficient for \(x^2\) is negative, use the same factoring process shown above. For example, complete the square for \(3x-x^2\) as shown below.

\(3x-x^2=-(x^2-3x)= -\left [ x^2-3x+ \left ( \frac{3}{2} \right )^2- \left ( \frac{3}{2} \right )^2\right ]= \left ( \frac{3}{2} \right )^2- \left (x-\frac{3}{2} \right )^2\)

Example 5.7.5 Completing the Square with Square Root

The region bounded by the graph for \(f\), the \(x\)-axis, \(x=\frac{3}{2}\), and \(x=\frac{9}{4}\) is \(\pi/6\).
Figure 5.7.1

Find the area for the region bounded by the graph for

$$f(x)=\frac{1}{\sqrt{3x-x^2}}$$

Solution The graph and area are shown in Figure 5.7.1.

$$\text{Area}$$

$$=\int_{3/2}^{9/4} \frac{1}{\sqrt{3x-x^2}}\:dx$$

$$=\int_{3/2}^{9/4} \frac{dx}{\sqrt{(3/2)^2-[x-(3/2)]^2}}\:dx$$

     Use completed square form derived above.

$$= \arcsin \left. \frac{x-(3/2)}{3/2} \right ]_{3/2}^{9/4}$$

$$= \arcsin \frac{1}{2}- \arcsin 0$$

$$=\frac{\pi}{6}$$

\(\approx 0.524\)

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Basic Integration Rules Review

The introduction to the basic integration rules is complete. Practice each rule enough so they are committed to memory.

Basic Integration Rules \((a>0\)

Calculus I 05.07.02.png

The discussion has covered all the rules to differentiate any elementary function. For integration, there are plenty more to learn.

The integration rules listed above are primarily those that were happened on while discussing differentiation rules. Rules for finding the antiderivative for a general product or quotient, the natural logarithmic function, or the inverse trigonometric functions have not been covered. More importantly, the rules require a proper \(du\) corresponding to the \(u\) in the formula, or they do not work. More on integration techniques is needed, which the discussion covers in Chapter 8. Examples 5.7.6 and 5.7.7 should give a better feeling for the integration problems that can, and cannot solve with the techniques and rules discussed so far.

Example 5.7.6 Comparing Integration Problems

Find all the possible integrals using formulas and techniques discussed so far.

$$\textbf{a.}\: \int \frac{dx}{x\sqrt{x^2-1}}$$

$$\textbf{b.}\: \int \frac{x\:dx}{\sqrt{x^2-1}}$$

$$\textbf{c.}\: \int \frac{dx}{\sqrt{x^2-1}}$$

Solution a. This one fits the Arcsecant Rule so it can be integrated.

$$\int \frac{dx}{x\sqrt{x^2-1}} = \operatorname{arcsec} \left | x \right |+C$$

b. This one fits the Power Rule, so it can be integrated.

$$\textbf{b.}\: \int \frac{x\:dx}{\sqrt{x^2-1}}$$

$$=\frac{1}{2} \int (x^2-1)^{-1/2} (2x)\:dx$$

$$=\frac{1}{2} \left [ \frac{(x^2-1)^{1/2} }{1/2} \right ] +C$$

\(=\sqrt{x^2-1}+C\)

c. This one cannot be integrated using the techniques discussed thus far. Verify this by trying the techniques summarized above.

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Example 5.7.7 Comparing Integration Problems

Find all the possible integrals using formulas and techniques discussed so far.

$$\textbf{a.}\: \int \frac{dx}{x \ln x}$$

$$\textbf{b.}\: \int \frac{ \ln x\:dx}{x}$$

$$\textbf{c.}\: \int \ln x \: dx$$

Solution a. This one can be integrated because it fits the Log Rule.

$$\int \frac{dx}{x \ln x}$$

$$= \int \frac{1/x}{\ln x}\:dx$$

$$=\ln \left | \ln x \right | +C$$

b. This one can be integrated because it fits the Power Rule.

$$\int \frac{ \ln x\:dx}{x}$$

$$=\int \left ( \frac{1}{x} \right ) \left ( \ln x \right )^1\:dx$$

$$=\frac{\left ( \ln x \right )^2}{2}+C$$

c. This one cannot be integrated using the techniques discussed so far.

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Internal Links

Parent Article: Calculus I 05 Logarithmic, Exponential, and Other Transcendental Functions