CHAPTER 12 Techniques of Integration I: Substitution, Powers of Trig Functions

1. Substitution

By a technique of integration we mean a method which can be used to find the anti-derivative (indefinite integral) of certain types of functions. We first review u-substitutions. We shall use an example to illustrate the thought process that goes into deciding on a substitution of the form u = u(x).

Example 12.1. Find the x-coordinate of the center of mass of the uniform plate that is bounded by the x- and y-axes, the line x = 10, and the curve y = 1/(1 + 0.01x^{2}).

We must compute \overline{x} = ∫_{0}^{10} x dx/(1 + 0.01x^{2}) / ∫_{0}^{10} dx/(1 + 0.01x^{2}). At first the

top and bottom integrals look similar: both have (1 + 0.01x^{2}) in the denominator. However, they are actually quite different.

In the bottom we have something very close to ∫ du/(1+u^{2}) =Arctan u+C. The only problem is the constant 0.01 in front of x^{2}. So we choose u so that u^{2} = 0.01x^{2}, i.e., u = 0.1x. Then x = 10u, dx = 10du. The limits of integration also change: x = 0 ⟹ u = 0, and x = 10 ⟹ u = 0.1 · 10 = 1. Hence, the bottom integral becomes ∫_{0}^{1} 10 du/(1 + u^{2}) = 10 Arctan u |_{0}^{1}= 10(π/4) = 2.5π.

The top integral, on the other hand, has nothing to do with Arctan. The x next to the dx changes everything. Namely, x is essentially the derivative of the denominator. Thus, a substitution u =denominator will lead to something like ∫ du/u = ln |u| + C. More precisely, we set u = 1 + 0.01x^{2}, so that du = 0.02x dx, and hence x dx = 50 du. This gives us 50 du/u inside the integral. The limits of integration change as follows: x = 0 ⟹ u = 1 + 0.01 · 0 = 1, and x = 10 ⟹ u = 1 + 0.01 · 10^{2} = 2. Thus, the top integral becomes 50 ∫_{1}^{2} du/u = 50 ln 2. Finally, we compute \overline{x} = 50 ln 2/2.5π = 4.4127.

2. Odd Powers of Sine and Cosine

The technique of u-substitution can be used to evaluate any integral of the form

∫ sin^{n}(ax + b) dx or ∫ cos^{n}(ax + b) dx, where n is a positive odd number.

The method works as follows. Say we want to integrate sin^{n}. First take out one sin and put it with the dx. What remains will be sin to an even power. Express this in terms of cos (using sin^{2} = 1 − cos^{2}). Now make a u-substitution for the cos. You will end up with only a polynomial to integrate. (If we had cos^{n} to integrate, we would do the same thing, with the roles of sin and cos reversed.) We illustrate with an example.

Example 12.2. Find ∫_{0}^{1} sin^{5}(πx)dx.

We first write

∫_{0}^{1} sin^{5}(πx)dx = ∫_{0}^{1} sin^{4}(πx) sin(πx)dx = ∫_{0}^{1}(1 − cos^{2}(πx))^{2} sin(πx)dx.

Next, we make the substitution u = cos(πx), so that du = −π sin(πx). Then sin(πx)dx = −\frac{1}{π} du, and so our integral becomes −\frac{1}{π} ∫ (1 − u^{2})^{2}du. The limits of integration change as follows: x = 0 ⟹ u = cos 0 = 1, and x = 1 ⟹ u = cos π = −1. To integrate the polynomial, we expand the square: (1 − u^{2})^{2} = 1 − 2u^{2} + u^{4}. So we obtain:

−\frac{1}{π} ∫_{1}^{−1} 1 − 2u^{2} + u^{4} du = −\frac{1}{π} (u − \frac{2}{3}u^{3} + \frac{1}{5}u^{5}) |_{1}^{−1}= \frac{16}{15π} .

It is also possible to use this method to integrate negative odd powers of sine and cosine. However, in that case instead of a polynomial in u you end up with a rational function (i.e., you have a denominator with a polynomial in it). We will treat some cases of this in part III of techniques of integration.

3. Even Powers of Sine and Cosine

The above technique does not work if you want to integrate something like cos^{2} or sin^{4}. To handle even powers one needs a trig identity, namely, the cos(2A) formula: cos(2A) = cos^{2} A−sin^{2} A. When combined with the trig identity 1 = cos^{2} A+sin^{2} A, this gives formulas for either cos^{2} A or sin^{2} A in terms of cos(2A). If we add these two identities and divide by 2, we get cos^{2} A; if we subtract them and divide by 2, we get sin^{2} A:

cos^{2} A + sin^{2} A = 1  cos^{2} A + sin^{2} A = 1
+ \underline{cos^{2} A − sin^{2} A} = \underline{cos(2A)}  − \underline{cos^{2} A − sin^{2} A} = \underline{cos(2A)}
2 cos^{2} A = 1 + cos(2A)  2 sin^{2} A = 1 − cos(2A) ,

i.e.,

cos^{2} A = \frac{1 + cos(2A)}{2},  sin^{2} A = \frac{1 − cos(2A)}{2}.

We now illustrate the use of these two identities in integration problems.

Example 12.3. Find the average value of sin^{2} x between 0 and π.

The average value of a function f(x) on the interval [a, b] is the integral of the function divided by the length of the interval: \frac{1}{b−a} ∫_{a}^{b} f(x)dx. So the average value of sin^{2} x is

\frac{1}{π} ∫_{0}^{π} sin^{2} x dx = \frac{1}{π} ∫_{0}^{π} (\frac{1 − cos(2x)}{2}) dx

= \frac{1}{π} (\frac{1}{2}x − \frac{1}{2} · \frac{1}{2} sin(2x)) |_{0}^{π}

= \frac{1}{π} (\frac{π}{2} − \frac{1}{4}(0 − 0)) = \frac{1}{2}.

Example 12.4. Find the anti-derivative of cos^{4} x.

We have

∫ cos^{4} x dx = ∫ (cos^{2} x)^{2} dx = ∫ (\frac{1 + cos(2x)}{2})^{2} dx

= ∫ (\frac{1 + 2 cos(2x) + cos^{2}(2x)}{4}) dx.

We now use the identity for cos^{2}(A) once more, this time with A = 2x: cos^{2}(2x) = (1+cos(4x))/2. In this way we end up replacing all powers of cosine with expressions of the form cos(ax), which are easy to integrate. After writing

1 + 2 cos(2x) + cos^{2}(2x) = 1 + 2 cos(2x) + (1 + cos(4x))/2

= \frac{3}{2} + 2 cos(2x) + \frac{1}{2} cos(4x)

and dividing this by 4, we obtain

∫ cos^{4} x dx = ∫ (\frac{3}{8} + \frac{1}{2} cos(2x) + \frac{1}{8} cos(4x)) dx

= \frac{3}{8}x + \frac{1}{4} sin(2x) + \frac{1}{32} sin(4x) + C.

Homework 1.

1. Find the x-coordinate of the center of mass of the flat plate bounded by the xand y-axes, the line x = 5, and the curve y = 1/\sqrt{1 − 0.01x^{2}}.

2. Find the anti-derivative of (a) sin(2πt)/\sqrt{2 + cos(2πt)};

(b) cos(ax + b)/ sin^{8}(ax + b).

3. Find the volume of the solid obtained by revolving y = cos x around the x-axis between x = 0 and x = π/4.

4. Find the average value of sin^{n}x between 0 and π/2 when n is (a) 3, (b) 4, (c) 5, (d) 6, (e) 7.