The previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral.

Rewrite \(\tan(x)\) as \(\sin(x) /\cos(x)\text{.}\) While the presence of a composition of functions may not be immediately obvious, recognize that \(\cos(x)\) is “inside” the \(1/x\) function. Therefore, we see if setting \(u = \cos(x)\) returns usable results. We have that \(du = -\sin(x) \ dx\text{,}\) hence \(-du = \sin(x) \ dx\text{.}\) We can integrate: \begin{align*} \int \tan(x) \ dx \amp = \int \frac{\sin(x) }{\cos(x) }\ dx\\ \amp = \int \frac1{\underbrace{\cos(x) }_u}\underbrace{\sin(x) \ dx}_{-du}\\ \amp = \int \frac {-1}u \ du\\ \amp = -\ln\abs{u} + C\\ \amp = -\ln\abs{\cos(x) } + C. \end{align*}

Some texts prefer to bring the \(-1\) inside the logarithm as a power of \(\cos(x)\text{,}\) as in: \begin{align*} -\ln\abs{\cos(x) } + C \amp = \ln\abs{(\cos(x) )^{-1}} + C\\ \amp = \ln\abs{ \frac{1}{\cos(x) }} + C\\ \amp = \ln\abs{\sec(x) } + C. \end{align*}

Thus the result they give is \(\int \tan(x) \ dx = \ln\abs{\sec(x) } + C\text{.}\) These two answers are equivalent.

This example employs a wonderful trick: multiply the integrand by “1” so that we see how to integrate more clearly. In this case, we write “1” as \begin{equation*} 1 = \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) }. \end{equation*}

This may seem like it came out of left field, but it works beautifully. Consider: \begin{align*} \int \sec(x) \ dx \amp = \int \sec(x) \cdot \frac{\sec(x) + \tan(x) }{\sec(x) + \tan(x) }\ dx\\ \amp = \int \frac{\sec^2(x) + \sec(x) \tan(x) }{\sec(x) + \tan(x) }\ dx.\\ \end{align*} Now let \(u = \sec(x) +\tan(x)\text{;}\) this means \(du = (\sec(x) \tan(x) + \sec^2(x) )\ dx\text{,}\) which is our numerator. Thus: \begin{align*} \amp = \int \frac{du}{u}\\ \amp = \ln\abs{u} + C\\ \amp = \ln\abs{\sec(x) +\tan(x) } + C. \end{align*}

We have a composition of functions as \(\cos^2(x) = \big(\cos(x) \big)^2\text{.}\) However, setting \(u = \cos(x)\) means \(du = -\sin(x) \ dx\text{,}\) which we do not have in the integral. Another technique is needed.

The process we'll employ is to use a Power Reducing formula for \(\cos^2(x)\) (perhaps consult the back of this text for this formula), which states \begin{equation*} \cos^2(x) = \frac{1+\cos(2x)}{2}. \end{equation*}

The right hand side of this equation is not difficult to integrate. We have: \begin{align*} \int \cos^2(x) \ dx \amp = \int \frac{1+\cos(2x)}2\ dx\\ \amp = \int \left( \frac12 + \frac12\cos(2x)\right)\ dx.\\ \end{align*} Now use Key Idea 6.1.4: \begin{align*} \amp = \frac12x + \frac12\frac{\sin(2x)}{2} + C\\ \amp = \frac12x + \frac{\sin(2x)}4 + C. \end{align*}

We'll make significant use of this power–reducing technique in future sections.

One may try to start by setting \(u\) equal to either the numerator or denominator; in each instance, the result is not workable.

When dealing with rational functions (i.e., quotients made up of polynomial functions), it is an almost universal rule that everything works better when the degree of the numerator is less than the degree of the denominator. Hence we use polynomial division.

We skip the specifics of the steps, but note that when \(x^2+2x+1\) is divided into \(x^3+4x^2+8x+5\text{,}\) it goes in \(x+2\) times with a remainder of \(3x+3\text{.}\) (You can get this same result by using the expand() command in a computer algebra system). Thus \begin{equation*} \frac{x^3+4x^2+8x+5}{x^2+2x+1} = x+2 + \frac{3x+3}{x^2+2x+1}. \end{equation*}

Integrating \(x+2\) is simple. The fraction can be integrated by setting \(u = x^2+2x+1\text{,}\) giving \(du = (2x+2)\ dx\text{.}\) This is very similar to the numerator. Note that \(du/2 = (x+1)\ dx\) and then consider the following: \begin{align*} \int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\ dx \amp = \int \left(x+2 + \frac{3x+3}{x^2+2x+1}\right)\ dx\\ \amp = \int (x+2)\ dx + \int \frac{3(x+1)}{x^2+2x+1}\ dx\\ \amp = \frac12x^2+2x+C_1 + \int \frac{3}{u}\frac{du}{2}\\ \amp = \frac12x^2+2x+C_1 + \frac32\ln\abs{u} + C_2\\ \amp = \frac12x^2+2x+\frac32\ln\abs{x^2+2x+1} + C. \end{align*}

In some ways, we “lucked out” in that after dividing, substitution was able to be done. In later sections we'll develop techniques for handling rational functions where substitution is not directly feasible.

We already know how to integrate this particular example. Rewrite \(\sqrt{x}\) as \(x^\frac12\) and simplify the fraction: \begin{equation*} \frac{x^2+2x+3}{x^{1/2}} = x^\frac32 + 2x^\frac12 + 3x^{-\frac12}. \end{equation*}

We can now integrate using the Power Rule: \begin{align*} \int \frac{x^2+2x+3}{x^{1/2}}\ dx \amp = \int\left(x^\frac32 + 2x^\frac12 + 3x^{-\frac12}\right)\ dx\\ \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12 + C \end{align*}

This is a perfectly fine approach. We demonstrate how this can also be solved using substitution as its implementation is rather clever.

Let \(u = \sqrt{x} = x^\frac12\text{;}\) therefore \begin{equation*} du = \frac{1}{2\sqrt{x}}\ dx \Rightarrow 2du = \frac{1}{\sqrt{x}}\ dx \text{.} \end{equation*}

This gives us \(\ds \int \frac{x^2+2x+3}{\sqrt{x}}\ dx = \int (x^2+2x+3)\cdot2\ du\text{.}\) What are we to do with the other \(x\) terms? Since \(u = x^\frac12\text{,}\) \(u^2 = x\text{,}\) etc. We can then replace \(x^2\) and \(x\) with appropriate powers of \(u\text{.}\) We thus have \begin{align*} \int \frac{x^2+2x+3}{\sqrt{x}}\ dx \amp = \int (x^2+2x+3)\cdot2\ du\\ \amp = \int 2(u^4 + 2u^2 + 3)\ du\\ \amp = \frac25u^5 + \frac43u^3 + 6u + C\\ \amp = \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12+C \end{align*} which is obviously the same answer we obtained before. In this situation, substitution is arguably more work than our other method. The fantastic thing is that it works. It demonstrates how flexible integration is.

The integrand looks similar to the derivative of the arctangent function. Note: \begin{align*} \frac{1}{25+x^2} \amp = \frac{1}{25\left(1+\frac{x^2}{25}\right)}\\ \amp = \frac{1}{25(1+\left(\frac{x}{5}\right)^2)}\\ \amp = \frac{1}{25}\frac{1}{1+\left(\frac{x}{5}\right)^2}\ \text{.} \end{align*}

Thus \begin{equation*} \int\frac{1}{25+x^2}\ dx = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\ dx. \end{equation*}

This can be integrated using Substitution. Set \(u = x/5\text{,}\) hence \(du = dx/5\) or \(dx=5\ du\text{.}\) Thus \begin{align*} \int\frac{1}{25+x^2}\ dx \amp = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\ dx\\ \amp = \frac15\int \frac{1}{1+u^2}\ du\\ \amp = \frac15\tan^{-1}(u) + C\\ \amp = \frac15\tan^{-1}\left(\frac x5\right)+C \end{align*}

Each can be answered using a straightforward application of Theorem 6.1.16.

\(\ds \int \frac{1}{9+x^2}\ dx = \frac13\tan^{-1}\left(\frac x3\right) + C\text{,}\) as \(a = 3\text{.}\)\(\ds \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\ dx = 10\sec^{-1}(10x) + C\text{,}\) as \(a = \frac1{10}\text{.}\)\(\ds \int \frac{1}{\sqrt{5-x^2}} = \sin^{-1}\left(\frac{x}{\sqrt{5}}\right)+C\text{,}\) as \(a = \sqrt{5}\text{.}\)

Initially, this integral seems to have nothing in common with the integrals in Theorem 6.1.16. As it lacks a square root, it almost certainly is not related to arcsine or arcsecant. It is, however, related to the arctangent function.

We see this by completing the square in the denominator. We give a brief reminder of the process here.

Start with a quadratic with a leading coefficient of 1. It will have the form of \(x^2 + bx + c\text{.}\) Take 1/2 of \(b\text{,}\) square it, and add/subtract it back into the expression. I.e., \begin{align*} x^2+bx+ c \amp = \underbrace{x^2 + bx + \frac{b^2}4}_{(x+b/2)^2} - \frac{b^2}4 + c\\ \amp = \left(x+\frac b2\right)^2 + c-\frac{b^2}4 \end{align*}

In our example, we take half of \(-4\) and square it, getting \(4\text{.}\) We add/subtract it into the denominator as follows: \begin{align*} \frac{1}{x^2-4x+13} \amp = \frac{1}{\underbrace{x^2-4x+4}_{(x-2)^2}-4+13}\\ \amp =\frac{1}{(x-2)^2 + 9} \end{align*}

We can now integrate this using the arctangent rule. Technically, we need to substitute first with \(u=x-2\text{,}\) but we can employ Key Idea 6.1.4 instead. Thus we have \begin{align*} \int \frac{1}{x^2-4x+13}\ dx \amp = \int \frac{1}{(x-2)^2+9}\ dx \\ \amp = \frac13\tan^{-1}\frac{x-2}{3}+C \end{align*}

This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral: \begin{equation*} \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\ dx. \end{equation*}

We handle each separately. The first integral is handled using a straightforward application of Theorem 6.1.16:

\(\ds \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\sin^{-1}\frac{x}{4} + C\text{.}\)

The second integral is handled by substitution, with \(u = 16-x^2\text{.}\) \(\ds \int\frac{x}{\sqrt{16-x^2}}\ dx\text{:}\) Set \(u = 16-x^2\text{,}\) so \(du = -2x\ dx\) and \(x\ dx = -du/2\text{.}\) We have \begin{align*} \int\frac{x}{\sqrt{16-x^2}}\ dx \amp = \int\frac{-du/2}{\sqrt{u}}\\ \amp = -\frac12\int \frac{1}{\sqrt{u}}\ du\\ \amp = - \sqrt{u} + C\\ \amp = -\sqrt{16-x^2} + C. \end{align*}

Combining these together, we have \begin{equation*} \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\sin^{-1}\frac x4 + \sqrt{16-x^2}+C. \end{equation*} As with all definite integrals, you can check your work by differentiation.

Observing the composition of functions, let \(u=3x-1\text{,}\) hence \(du = 3\ dx\text{.}\) As \(3\ dx\) does not appear in the integrand, divide the latter equation by 3 to get \(du/3 = dx\text{.}\)

By setting \(u = 3x-1\text{,}\) we are implicitly stating that \(g(x) = 3x-1\text{.}\) Theorem 6.1.20 states that the new lower bound is \(g(0) = -1\text{;}\) the new upper bound is \(g(2) = 5\text{.}\) We now evaluate the definite integral: \begin{align*} \int_1^2 \cos(3x-1) \ dx \amp = \int_{-1}^5 \cos(u) \frac{du}{3}\\ \amp = \frac{1}{3} \sin(u) \Big|_{-1}^5\\ \amp = \frac{1}{3}\big(\sin(5) - \sin(-1)\big)\\ \amp \approx -0.039. \end{align*}

Notice how once we converted the integral to be in terms of \(u\text{,}\) we never went back to using \(x\text{.}\)

The graphs in Figure 6.1.22 tell more of the story. In Figure 6.1.22.(a) the area defined by the original integrand is shaded, whereas in Figure 6.1.22.(b) the area defined by the new integrand is shaded. In this particular situation, the areas look very similar; the new region is “shorter” but “wider,” giving the same area.

We saw the corresponding indefinite integral in Example 6.1.6. In that example we set \(u = \sin(x)\) but stated that we could have let \(u = \cos(x)\text{.}\) For variety, we do the latter here.

Let \(u = g(x) = \cos(x)\text{,}\) giving \(du = -\sin(x) \ dx\) and hence \(\sin(x) \ dx = -du\text{.}\) The new upper bound is \(g(\pi/2) = 0\text{;}\) the new lower bound is \(g(0) = 1\text{.}\) Note how the lower bound is actually larger than the upper bound now. We have \begin{align*} \int_0^{\pi/2} \sin(x) \cos(x) \ dx \amp = \int_1^0 -u\ du \quad \text{(switch bounds and change sign) }\\ \amp = \int_0^1 u\ du\\ \amp = \frac12u^2\Big|_0^1= 1/2. \end{align*}

In Figure 6.1.24 we have again graphed the two regions defined by our definite integrals. Unlike the previous example, they bear no resemblance to each other. However, Theorem 6.1.20 guarantees that they have the same area.