3. Derivatives of Trigonometric Functions | Section 4 Exercises | Trigonometric Functions Main Page | "RealWorld" Page | Everything for Calculus | Español |
Recall from the definition of an antiderivative that, if
then
That is, every time we have a differentiation formula, we get an integration formula for nothing. Here is a list of some of them.
$\frac{d}{dx} \sin x = \cos x$ | $\int \cos x\ dx = \sin x + C$ |
$\frac{d}{dx} \cos x = -\sin x$ | $\int \sin x\ dx = -\cos x + C$ |
$\frac{d}{dx} \tan x = \sec^2x$ | $\int \sec^2x\ dx = \tan x + C$ |
$\frac{d}{dx} \cotan x = -\cosec^2 x$ | $\int \cosec^2x\ dx = -\cotan x + C$ |
$\frac{d}{dx} \sec x = \sec x \tan x$ | $\int (\sec x \tan x) dx = \sec x + C$ |
$\frac{d}{dx} \cosec x = -\cosec x \cotan x$ | $\int (\cosec x \cotan x) dx = -\cosec x + C$ |
Question
What about the other four?
Answer
We shall obtain some of them below, and leave others to the exercise set. (Some of them have already appeared as derivatives in Exercise Set 3...).
Compute the following.
(a) | $\int (3\sin x - 4\sec^2x) dx$ | (b) | $\int \cos(2x - 6) dx$ |
(c) | $\int \sin x\ \cos^2x\ dx$ | (d) | $\int \tan x\ dx$ |
Solution
(a) Consulting the table above,
$\int (3\sin x - 4\sec^2x) dx$ | $= \int 3\sin x\ dx - \int 4\sec^2x\ dx$ | (properties of integrals) |
$= 3 \int \sin x\ dx - 4 \int \sec^2x\ dx$ | (properties of integrals) | |
$= -3\cos x - 4\tan x + C$ | (from the table) |
$\color{blue}{u = 2x-6}$
$\color{blue}{\frac{du}{dx}=2}$ $\color{blue}{dx = \frac{1}{2} du}$ |
We now have
$= \int \cos u \frac{1}{2} du$ | (using the substitution) | |
$= \frac{1}{2} \int \cos u\ du$ | (properties of integrals) | |
$= \frac{1}{2} \sin\ u + C$ | (from the table) | |
$= \frac{1}{2} \sin(2x-6) + C.$ | (using the substitution) |
$\color{blue}{u = \cos x }$
$\color{blue}{\frac{du}{dx} = -\sin x}$ $\color{blue}{dx = - \frac{1}{\sin x} du}$ |
We now have
$\int \sin x \cos^2x\ dx$ | $= \int (\sin x)u^2 \left(- \frac{1}{\sin x}\right) du$ | (using the substitution) |
$=- \int u^2 du$ | ||
$=- \frac{u^3}{3} + C$ | ||
$=- \frac{\cos^3x}{3} + C.$ | (using the substitution) |
$\int \tan x\ dx$ | $= \int \frac{\sin x}{\cos x} dx$ | |
$= \int \frac{\sin x}{u} \left(- \frac{1}{\sin x} \right) du$ | (using the substitution) | |
$= - \int \frac{1}{u} du$ | ||
$= - \ln \|u\| + C$ | ||
$= - \ln \|\cos x\| + C.$ | (using the substitution) |
Before we go on...
The method in part (b) gives us the following more general formulas:
$\int \cos x\ dx = \sin x + C$ | $\int \cos(ax+b) dx = \frac{1}{a} \sin(ax+b) + C$ |
$\int \sin x\ dx = -\cos x + C$ | $\int \sin(ax+b) dx = - \frac{1}{a} \cos(ax+b) + C$ |
Not to keep you in suspense, here are the antiderivatives of all six trigonometric functions. (You will obtain them in the exercises.)
$\int \cos x\ dx = \sin x + C$ | $\int \cos(ax+b) dx = \frac{1}{a} \sin(ax+b) + C$ |
$\int \sin x\ dx = -\cos x + C$ | $\int \sin(ax+b) dx = - \frac{1}{a} \cos(ax+b) + C$ |
$\int \tan x\ dx = - \ln \|\cos x\| + C$ | $\int \tan(ax+b) dx = -\frac{1}{a} \ln \|\cos(ax+b)\| + C$ |
$\int \cotan x\ dx = \ln \|\sin x\| + C$ | $\int \cotan(ax+b) dx = \frac{1}{a} \ln \|\sin(ax+b)\| + C$ |
$\int \sec x\ dx = \ln \|\sec x + \tan x\| + C$ | $\int \sec(ax+b) dx = \frac{1}{a} \ln \|\sec(ax+b) + \tan(ax+b)\| + C$ |
$\int \cosec x\ dx = -\ln \|\cosec x + \cotan x\| + C$ | $\int \cosec(ax+b) dx = - \frac{1}{a} \ln \|\cosec(ax+b) + \cotan(ax+b)\| + C$ |
Solution
Since total sales are given by the definite integral of monthly sales for the given period ($t = 2$ to $t = 6$), we have, consulting the above table,
Total Sales | $=$ |
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$=$ |
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$=$ |
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We can also use the tabular method of integration by parts discussed in Section 7.1 of Calculus Applied to the Real World, or Section 14.1 of Finite Mathematics and Calculus Applied to the Real World.
Example 3Evaluate the following integrals
Solution
(a) Since repeated differentiation of the first term $(3x^2-2x+1)$ results in zero, we place it in the "D" column:
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This gives:
(b) Repeated differentiation does not annihilate either term. Actually, it doesn't matter which term we place in the "D" column, so let us place the trigonometric function there:
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This gives
(We will add the constant of integration after we are done.) Notice that we have ended up with the same integral on the right as the one we started with. Calling this integral I gives:
and we can now solve for I:
that is,
so that
3. Derivatives of Trigonometric Functions | Section 4 Exercises | Trigonometric Functions Main Page | "RealWorld" Page | Everything for Calculus | Español |
Mail us at:
Stefan Waner (matszw@hofstra.edu) | Steven R. Costenoble (matsrc@hofstra.edu) |