Exercise 7.1 · Q1: An anti derivative of $\sin 2x$

Basics you need

Solution

1. Make a first guess from a known derivative

   We need a function whose derivative is a sine. The standard formulae give one immediately:

   $$\frac{d}{dx}(-\cos x) = \sin x.$$

   Our function has $2x$ in place of $x$⁠, so the natural first guess is $-\cos 2x$⁠.

   Inspection always starts from a derivative we already know. The only derivative that produces a plain sine is that of $-\cos$⁠, so the answer must be built from $-\cos 2x$⁠. The guess is only a candidate until we differentiate it.
   Item (iii) of the list of standard formulae, Section 7.2: $\frac{d}{dx}(-\cos x) = \sin x$⁠, equivalently $\int \sin x\,dx = -\cos x + C$⁠. Basics: trigonometric facts for integrals

2. Differentiate the guess (chain rule)

   Write the guess as an outer function of an inner one: inner $u = 2x$⁠, outer $-\cos u$⁠. Then

   $$\begin{aligned} \frac{d}{dx}(-\cos 2x) &= \sin u \cdot \frac{du}{dx} \\ &= \sin 2x \cdot 2 \\ &= 2\sin 2x. \end{aligned}$$

   The guess gives $2\sin 2x$⁠: the right function, but twice too big.

   The derivative of the outer $-\cos u$ with respect to $u$ is $\sin u$⁠, and the chain rule multiplies it by $\frac{du}{dx} = 2$⁠. That extra factor $2$ comes from the inner $2x$⁠: the curve $-\cos 2x$ runs through its values twice as fast as $-\cos x$⁠, so its slopes are twice as steep.
   Chain rule (Theorem 4, Section 5.3.1 of Chapter 5): $\frac{d}{dx}g(u(x)) = g'(u(x))\,u'(x)$⁠. Basics: the chain rule backwards

3. Remove the extra factor 2

   Multiplying a function by a constant multiplies its derivative by the same constant. So multiply the guess by $\frac12$⁠:

   $$\begin{aligned} \frac{d}{dx}\left(-\tfrac12\cos 2x\right) &= \tfrac12\cdot 2\sin 2x \\ &= \sin 2x. \end{aligned}$$

   Hence $-\frac12\cos 2x$ is an anti derivative of $\sin 2x$.

   The constant factor rule $(kF)' = kF'$ lets us scale the guess freely. We scale by exactly the reciprocal of the unwanted factor, $\frac12$⁠, so that $\frac12 \cdot 2 = 1$ and nothing but $\sin 2x$ is left. This is the same move the book makes for $\cos 2x$ in Example 1 (i).
   Constant multiple rule for derivatives (Section 5.3): $\frac{d}{dx}(kF) = k\frac{dF}{dx}$⁠. Example 1

4. Check: differentiate the answer back

   Let $F(x) = -\frac12\cos 2x$⁠. With inner $u = 2x$ (so $u' = 2$⁠) and $\frac{d}{du}\cos u = -\sin u$⁠:

   $$\begin{aligned} F'(x) &= -\tfrac12 \cdot (-\sin 2x) \cdot 2 \\ &= \sin 2x. \;\checkmark \end{aligned}$$

   The derivative is exactly the given function, for every real $x$⁠.

   Inspection is guessing, and a guess is only an answer once it has been checked. Differentiating back is the definition of "anti derivative" applied directly: if $F' = f$⁠, then $F$ is an anti derivative of $f$⁠, with nothing left to doubt.
   Property (I), Section 7.2.1: differentiation and integration undo each other. Basics: integration undoes differentiation

5. All anti derivatives: add the constant

   Adding any constant does not change the derivative, so every function $-\frac12\cos 2x + C$ is an anti derivative too, and by the Remark in Section 7.2 these are all of them:

   $$\int \sin 2x\,dx = -\tfrac12\cos 2x + C.$$
   The question asks for an anti derivative, so $-\frac12\cos 2x$ alone answers it. But the integral sign means the whole family, and the picture shows why: lifting a graph up or down changes no slope, and two functions with the same derivative on an interval differ only by a constant.
   Remark in Section 7.2: functions with the same derivative on an interval differ by a constant, so $\{F + C : C \in \R\}$ is the family of all anti derivatives. Lesson: integration as anti differentiation

Every fact used, and where it comes from