7.5: Integrals Involving Inverse Trigonometric Functions

Aside from the Museum Problem and its sporting variations introduced in the previous section, the primary use of the inverse trigonometric functions in calculus involves their role as antiderivatives of rational and algebraic functions. Each of the six differentiation patterns from the previous section provides us with an integral formula, but they give rise to only three essentially different patterns:
\begin{align*}
\int \frac{1}{\sqrt{1-x^2}} \ dx & = \arcsin(x) + C \qquad (\mbox{valid for: }-1 < x < 1)\\
\int \frac{1}{1+x^2} \ dx & = \arctan(x) + C\qquad (\mbox{valid for: }-\infty < x < \infty)\\
\int \frac{1}{\left|x\right|\sqrt{x^2-1}} \ dx & = \mbox{arcsec}(x) + C\qquad (\mbox{valid for: }\left|x\right| > 1)
\end{align*}
Most of the related antiderivative patterns you will need in practice arise from variations of these basic ones. Typically, you need to transform an integrand so that it exactly matches one of the basic patterns. (Why are the derivative patterns for \(\arccos\), \(\mbox{arccot}\) and \(\mbox{arccsc}\) of little use to us when finding antiderivatives of algebraic functions?)

Example \(\PageIndex{1}\)

Evaluate \(\displaystyle \int \frac{1}{16+x^2} \ dx\).

Solution

We can transform this integrand into the arctangent pattern by factoring \(16\) from the denominator\[\int \frac{1}{16+x^2} \ dx = \int \frac{1}{16\left(1+\frac{x^2}{16}\right)} \ dx = \frac{1}{16} \int \frac{1}{1+\left(\frac{x}{4}\right)^2} \ dx\nonumber\]and then using the substitution \(\displaystyle u = \frac{x}{4} \ \Rightarrow du = \frac14 \, dx \ \Rightarrow \ 4\, du = dx\):\[\frac{1}{16} \int \frac{1}{1+u^2} \cdot 4 \, du = \frac{4}{16} \int \frac{1}{1+u^2}\ du = \frac14 \arctan(u) + C\nonumber\]Replacing \(u\) with \(\displaystyle \frac{x}{4}\) we get \(\displaystyle \frac14 \arctan\left(\frac{x}{4}\right)+C\) as our final answer.

Practice \(\PageIndex{1}\)

Evaluate \(\displaystyle \int \frac{1}{1+9x^2}\ dx\) and \(\displaystyle \int \frac{1}{\sqrt{25-x^2}}\ dx\).

Answer

For the first integral, write:\[\int \frac{1}{1+9x^2} \ dx = \int \frac{1}{1+(3x)^2} \ dx\nonumber\]and substitute \(u = 3x \ \Rightarrow du = 3\, dx \ \Rightarrow \ \frac13 \, du = dx\) to get:\[\int \frac{1}{1+u^2} \cdot \frac13 \, du = \frac13 \int \frac{1}{1+u^2} \, du = \frac13 \arctan(u)+ C = \frac13\arctan(3x)+C\nonumber\]For the second integral, factor out \(25\) to get:\[\int \frac{1}{\sqrt{25-x^2}}\ dx = \int \frac{1}{5\sqrt{1-\frac{x^2}{25}}} \ dx = \frac15 \int \frac{1}{\sqrt{1-\left(\frac{x}{5}\right)^2}} \ dx\nonumber\]and then substitute \(\displaystyle u = \frac{x}{5} \ \Rightarrow \ du = \frac15 \, dx \ \Rightarrow \ 5\, du = dx\):\[\frac15 \int \frac{1}{\sqrt{1-\left(\frac{x}{5}\right)^2}}\cdot 5 \, dx = \int \frac{1}{\sqrt{1-u^2}}\, du = \arcsin(u) + K\nonumber\]Replacing \(u\) with \(\displaystyle \frac{x}{5}\) yields:\[\int \frac{1}{\sqrt{25-x^2}}\ dx = \arcsin\left(\frac{x}{5}\right) + K\nonumber\]

The integrands that arise most often contain patterns with the forms \(a^2 - x^2\), \(a^2 + x^2\) and \(x^2-a^2\), where \(a\) is some positive constant, so it is worthwhile to develop general integral patterns for these forms, list 
them in Appendix I, and refer to them when necessary:
\begin{align*}
\int \frac{1}{\sqrt{a^2-x^2}} \ dx & = \arcsin\left(\frac{x}{a}\right) + C \qquad (\mbox{valid for: }-a < x < a)\\
\int \frac{1}{a^2+x^2} \ dx & = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C \qquad (\mbox{valid for: }-\infty < x < \infty)\\
\int \frac{1}{\left|x\right|\sqrt{x^2-a^2}} \ dx & = \frac{1}{a} \mbox{arcsec}\left(\frac{x}{a}\right) + C \qquad (\mbox{valid for: }\left|x\right| > a)
\end{align*}
You can arrive at each of these general formulas by factoring the \(a^2\) out of the denominator and making a suitable change of variable (as in Example 1, with \(a\) in place of \(4\)). You can then check the result by differentiating. The \(\arctan\) pattern is, by far, the most common. The \(\arcsin\) pattern appears occasionally, and the \(\mbox{arcsec}\) pattern only rarely.

Example \(\PageIndex{2}\)

Develop the general formula for \(\displaystyle \int \frac{1}{\sqrt{a^2-x^2}}\ dx\) from the known formula for \(\displaystyle \int \frac{1}{\sqrt{1-x^2}} \ dx\). (Assume that \(a > 0\).)

Solution

Using Example 1 as a guide, factor \(a^2\) out of the denominator:\[\int \frac{1}{\sqrt{a^2-x^2}}\ dx = \int \frac{1}{a\sqrt{1-\frac{x^2}{a^2}}}\ dx = \frac{1}{a} \int \frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^2}} \ dx\nonumber\]Now substitute \(\displaystyle u = \frac{x}{a} \ \Rightarrow du = \frac{1}{a} \, dx \ \Rightarrow \ a\, du = dx\) to get:\[\frac{1}{a} \int \frac{1}{\sqrt{1-u^2}} \cdot a \, du = \frac{a}{a} \int \frac{1}{\sqrt{1-u^2}} \ du = \arcsin(u) + C\nonumber\]and replace \(u\) with \(\displaystyle \frac{x}{a}\) to get \(\displaystyle \arcsin\left(\frac{x}{a}\right) + C\), the desired result.

Practice \(\PageIndex{2}\)

Verify that the derivative of \(\displaystyle \frac{1}{a} \cdot \arctan\left(\frac{x}{a}\right)\) is \(\displaystyle \frac{1}{a^2+x^2}\).

Answer

Using the Chain Rule:\[\mbox{D}\left[\frac{1}{a}\arctan\left(\frac{x}{a}\right)\right] = \frac{1}{a} \cdot \frac{1}{1+\left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} = \frac{1}{a^2\left[1+\left(\frac{x}{a}\right)^2\right]} = \frac{1}{a^2+x^2}\nonumber\]

Example \(\PageIndex{3}\)

Evaluate \(\displaystyle \int \frac{1}{\sqrt{5-x^2}}\ dx\) and \(\displaystyle \int_1^3 \frac{1}{5+x^2} \ dx\).

Solution

The constant \(a\) needn't be an integer, so take \(a^2 = 5 \Rightarrow a = \sqrt{5}\):\[\int \frac{1}{\sqrt{5-x^2}}\ dx = \arcsin\left(\frac{x}{\sqrt{5}}\right) + C\nonumber\]using the pattern from Example 2, while the general \(\arctan\) pattern yields:
\begin{align*}\int_1^3 \frac{1}{5+x^2} \ dx &= \left[\frac{1}{\sqrt{5}} \arctan\left(\frac{x}{\sqrt{5}}\right)\right]_1^3 \\
&= \frac{1}{\sqrt{5}}\left[\arctan\left(\frac{3}{\sqrt{5}}\right) - \arctan\left(\frac{1}{\sqrt{5}}\right)\right]\end{align*}
or about \(0.228\).

The easiest way to integrate certain rational functions is to split the original integrand into two pieces.

Example \(\PageIndex{4}\)

Evaluate \(\displaystyle \int \frac{6x+7}{25+x^2}\ dx\).

Solution

The integrand splits nicely into the sum of two other functions that you can integrate more easily:\[\int \frac{6x+7}{25+x^2}\ dx = \int \frac{6x}{25+x^2}\ dx + \int \frac{7}{25+x^2}\ dx\nonumber\]In the first integral, use the subsitution \(u = 25+x^2 \Rightarrow du = 2x \, dx \Rightarrow \frac{1}{2}\, du = x \, dx\):\[\int \frac{6x}{25+x^2}\ dx = \frac{6}{2} \int \frac{1}{u} \, du = 3 \ln\left(\left|u\right|\right) + C_1 = 3\ln\left(25+x^2\right)+C_1\nonumber\]
Meanwhile, the second integral matches the general arctangent pattern with \(a = 5\):\[\int \frac{7}{25+x^2}\ dx = 7\int \frac{1}{5^2+x^2}\ dx = 7\cdot \frac15 \arctan\left(\frac{x}{5}\right) + C_2\nonumber\]Combining these two results yields:\[\int \frac{6x+7}{25+x^2}\ dx = \ln\left(25+x^2\right)^3 + \frac75 \arctan\left(\frac{x}{5}\right) + C\nonumber\]as our final answer. (The two constants \(C_1\) and \(C_2\) add up to another arbitrary constant, which we can simply call \(C\).)

Why can we remove the absolute value signs in the last step?

The antiderivative of a linear function divided by an irreducible quadratic polynomial will typically result in the sum of a logarithm and an arctangent.

Practice \(\PageIndex{3}\)

Evaluate \(\displaystyle \int \frac{4x+3}{x^2+7}\ dx\).

Answer

Split the integrand into two pieces:\[\int \frac{4x+3}{x^2+7} \ dx = \int \frac{4x}{x^2+7} \ dx + \int \frac{3}{x^2+7} \ dx\nonumber\]For the first integral, let \(u = x^2 + 7 \ \Rightarrow \ du = 2x \, dx \Rightarrow 2\, du = 4x\, dx\):\[\int \frac{4x}{x^2+7} \ dx = \int \frac{2}{u}\, du = 2\ln\left(\left|u\right|\right) + C_1 = 2\ln\left(x^2+7\right)+C_1\nonumber\]The second integral matches the arctangent pattern with \(a^2 = 7 \Rightarrow a = \sqrt{7}\):\[\int \frac{3}{x^2+7} \ dx = 3\int \frac{1}{\left(\sqrt{7}\right)^2 + x^2} \ dx = 3\cdot \frac{1}{\sqrt{7}} \arctan\left(\frac{x}{\sqrt{7}}\right)+C_2\nonumber\]Combining these results yields:\[\int \frac{4x+3}{x^2+7} \ dx = \ln\left(\left[x^2+7\right]^2\right) + \frac{3}{\sqrt{7}} \arctan\left(\frac{x}{\sqrt{7}}\right) + C\]

Problems

In Problems 1–24, evaluate the integral.

1. \(\displaystyle \int \frac{7}{\sqrt{9-x^2}} \ dx\)
2. \(\displaystyle \int \frac{9}{\sqrt{7-y^2}} \ dy\)
3. \(\displaystyle \int_0^1 \frac{3}{x^2+25} \ dx\)
4. \(\displaystyle \int_5^7 \frac{5}{x\sqrt{x^2-16}} \ dx\)
5. \(\displaystyle \int \frac{9}{\sqrt{49-x^2}} \ dx\)
6. \(\displaystyle \int_1^4 \frac{2}{7+x^2} \ dx\)
7. \(\displaystyle \int_6^{10} \frac{3}{x\sqrt{x^2-25}} \ dx\)
8. \(\displaystyle \int \frac{7}{(x-5)^2+9} \ dx\)
9. \(\displaystyle \int \frac{1}{(x-1)^2+1} \ dx\)
10. \(\displaystyle \int \frac{1}{x^2-2x+2} \ dx\)
11. \(\displaystyle \int_{-1}^1 \frac{e^x}{1+e^{2x}} \ dx\)
12. \(\displaystyle \int_{1}^e \frac{1}{x} \cdot \frac{3}{1+\left[\ln(x)\right]^2} \ dx\)
13. \(\displaystyle \int \frac{\cos(\theta)}{\sqrt{9-\sin^2(\theta)}} \ d\theta\)
14. \(\displaystyle \int \frac{8x}{16+x^2} \ dx\)
15. \(\displaystyle \int \frac{3x}{\sqrt{9+x^2}} \ dx\)
16. \(\displaystyle \int \frac{3x}{\sqrt{9-x^2}} \ dx\)
17. \(\displaystyle \int \frac{6x}{9+x^4} \ dx\)
18. \(\displaystyle \int \frac{6x}{\sqrt{9-x^4}} \ dx\)
19. \(\displaystyle \int \frac{1}{1+4x^2} \ dx\)
20. \(\displaystyle \int \frac{1}{\sqrt{1-9x^2}} \ dx\)
21. \(\displaystyle \int_0^{\infty} \frac{1}{3+x^2} \ dx\)
22. \(\displaystyle \int_0^2 \frac{1}{\sqrt{4-x^2}} \ dx\)
23. \(\displaystyle \int_0^{\sqrt{7}} \frac{1}{\sqrt{7-x^2}} \ dx\)
24. \(\displaystyle \int_0^{\infty} \frac{x}{1+x^4} \ dx\)

In Problems 25–28, solve the initial value problem.

25. \(\displaystyle \frac{dy}{dx} = \frac{y}{\sqrt{1-x^2}}\), \(y(0) = e\)
26. \(\displaystyle \frac{dy}{dx} = \frac{1}{y(1+x^2)}\), \(y(0) = 4\)
27. \(\displaystyle \frac{dy}{dx} = \frac{y^2}{9+x^2}\), \(y(1) = 2\)
28. \(\displaystyle \frac{dy}{dx} \cdot \sqrt{16-x^2} = y\), \(y(4) = 1\)

In Problems 29–32, evaluate the integral by splitting the integrand into two simpler functions.

29. \(\displaystyle \int \frac{8x+5}{x^2+9} \ dx\)
30. \(\displaystyle \int \frac{1-4x}{x^2+1} \ dx\)
31. \(\displaystyle \int \frac{7x+3}{x^2+10} \ dx\) 
32. \(\displaystyle \int \frac{x+5}{x^2+16} \ dx\) 

Problems 33–40 illustrate how we can sometimes decompose a difficult integral into simpler ones. (Hints: For 33, complete the square in the denominator; for 34, let \(u =\) denominator; for 35, write \(4x+20 = (4x+12)+8\).)

33. \(\displaystyle \int \frac{8}{x^2+6x+10} \ dx\)
34. \(\displaystyle \int \frac{4x+12}{x^2+6x+10} \ dx\)
35. \(\displaystyle \int \frac{4x+20}{x^2+6x+10} \ dx\)
36. \(\displaystyle \int \frac{7}{x^2+4x+5} \ dx\)
37. \(\displaystyle \int \frac{12x+24}{x^2+4x+5} \ dx\)
38. \(\displaystyle \int \frac{12x+31}{x^2+4x+5} \ dx\)
39. \(\displaystyle \int \frac{6x+15}{x^2+4x+20} \ dx\)
40. \(\displaystyle \int \frac{2x+5}{x^2-4x+13} \ dx\)