Exercise 7.2 · Q5: $\int \sin(ax+b)\cos(ax+b)\,dx$

Basics you need

Solution

1. Rewrite the product with the double-angle identity

   Put $\theta = ax+b$ in $\sin\theta\cos\theta = \frac12\sin 2\theta$⁠:

   $$\sin(ax+b)\cos(ax+b) = \tfrac12\sin\big(2(ax+b)\big).$$
   A product of two trigonometric functions has no standard formula, but a single sine of a linear expression does. The identity converts one into the other at no cost.
   $\sin 2\theta = 2\sin\theta\cos\theta$ (Class XI trigonometry). Basics: trigonometric identities

2. Substitute $t = 2(ax+b)$

   $$\begin{aligned} t &= 2ax + 2b, \\ \frac{dt}{dx} &= 2a, \\ dx &= \frac{dt}{2a}. \end{aligned}$$

   Therefore

   $$\begin{aligned} \int \tfrac12\sin\big(2(ax+b)\big)\,dx &= \tfrac12\int\sin t\cdot\frac{dt}{2a} \\ &= \frac{1}{4a}\int \sin t\,dt. \end{aligned}$$
   The inner expression $2ax+2b$ is linear, so its derivative $2a$ is a constant. We can divide by it because $a \ne 0$⁠. The constants $\frac12$ and $\frac{1}{2a}$ move outside the integral.
   Substitution, Section 7.3.1 (compare Example 5(i)); Property (IV), Section 7.2.1. Lesson: integration by substitution

3. Integrate $\sin t$

   $$\frac{1}{4a}\int\sin t\,dt = \frac{1}{4a}\left(-\cos t\right) + C = -\frac{\cos t}{4a} + C.$$
   Standard formula (iii) gives $\int\sin t\,dt = -\cos t + C$⁠. The constant $\frac{1}{4a}$ in front simply multiplies that result; a constant times an arbitrary constant is again written $C$⁠.
   Standard formula (iii), Section 7.2.

4. Return to $x$

   $$-\frac{\cos t}{4a} + C = -\frac{1}{4a}\cos 2(ax+b) + C.$$
   Replace $t$ by $2(ax+b)$⁠. This is valid for all real $x$ (with $a \ne 0$⁠).

5. Check: differentiate the answer

   Let $F(x) = -\frac{1}{4a}\cos(2ax+2b)$⁠. Chain rule with inner $u = 2ax+2b$ (so $u' = 2a$⁠) and outer $\cos u$⁠:

   $$\begin{aligned} F'(x) &= -\frac{1}{4a}\cdot\big(-\sin(2ax+2b)\big)\cdot 2a \\ &= \frac{2a}{4a}\,\sin(2ax+2b) \\ &= \tfrac12\sin\big(2(ax+b)\big) \\ &= \sin(ax+b)\cos(ax+b). \end{aligned}$$

   The last line uses the identity of Step 1 backwards. We are back at the integrand.

   Getting the integrand back is exactly what makes $F$ an antiderivative (Property (I), Section 7.2.1). The inner derivative $2a$ cancels the $a$ in $\frac{1}{4a}$⁠, which is why the method needs $a \ne 0$⁠.
   Chain rule, Chapter 5, Theorem 4; $(\cos u)' = -\sin u$⁠, Table 5.3. Basics: integration undoes differentiation

6. Another route gives a different-looking answer

   Instead, put $s = \sin(ax+b)$⁠. Then $ds = a\cos(ax+b)\,dx$⁠, and

   $$\begin{aligned} \int \sin(ax+b)\cos(ax+b)\,dx &= \frac1a\int s\,ds \\ &= \frac{s^2}{2a} + C' \\ &= \frac{\sin^2(ax+b)}{2a} + C'. \end{aligned}$$

   Using $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ with $\theta = ax+b$⁠:

   $$\frac{\sin^2(ax+b)}{2a} = \frac{1}{4a} - \frac{\cos 2(ax+b)}{4a}.$$

   So the two answers differ by the constant $\frac{1}{4a}$⁠, which is absorbed into the arbitrary constant. Both are correct.

   Indefinite integrals are families of functions. Two antiderivatives of the same function always differ by a constant (Property (II), Section 7.2.1), so different methods may give answers that look different. Showing the difference is a constant proves they agree.
   $\cos 2\theta = 1 - 2\sin^2\theta$⁠. Basics: trigonometric identities

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