Simple Integration of Trigonometric Functions (Part 2)

This lesson works through a set of simple integration problems, most of them involving trigonometric functions, and shows how to use a given condition to find the constant of integration.

Recovering a function from its derivative

We are given $f'(x) = 4x^3 - \dfrac{3}{x^4}$ with $f(2) = 0$, and we want $f(x)$. Since integrating the derivative gives back the function:

$f(x) = \int \left(4x^3 - 3x^{-4}\right)\, dx = 4\cdot\frac{x^4}{4} - 3\cdot\frac{x^{-3}}{-3} + C = x^4 + \frac{1}{x^3} + C$

Now apply $f(2) = 0$:

$2^4 + \frac{1}{2^3} + C = 0 \quad\Rightarrow\quad 16 + \frac{1}{8} + C = 0$

Taking the common denominator $8$ gives $C = -\dfrac{128 + 1}{8} = -\dfrac{129}{8}$, so

$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$

A polynomial, cosine and exponential mix

$\int \left(2x - 3\cos x + e^{x}\right)\, dx = 2\cdot\frac{x^2}{2} - 3\sin x + e^{x} + C = x^2 - 3\sin x + e^{x} + C$

Integral of secant times secant plus tangent

Multiply out first, then integrate term by term:

$\int \sec x\,(\sec x + \tan x)\, dx = \int \left(\sec^2 x + \sec x\,\tan x\right)\, dx = \tan x + \sec x + C$

This uses $\int \sec^2 x\,dx = \tan x$ and $\int \sec x\,\tan x\,dx = \sec x$.

Secant squared over cosecant squared

Write each function in terms of sine and cosine:

$\int \frac{\sec^2 x}{\csc^2 x}\, dx = \int \frac{1/\cos^2 x}{1/\sin^2 x}\, dx = \int \frac{\sin^2 x}{\cos^2 x}\, dx = \int \tan^2 x\, dx$

Using the identity $\tan^2 x = \sec^2 x - 1$:

$\int \left(\sec^2 x - 1\right)\, dx = \tan x - x + C$

Quotients with cosine squared denominators

First quotient. Split the fraction and rewrite each piece:

$\int \frac{2 - 3\sin x}{\cos^2 x}\, dx = \int \frac{2}{\cos^2 x}\, dx - \int \frac{3\sin x}{\cos^2 x}\, dx$

$= 2\int \sec^2 x\, dx - 3\int \sec x\,\tan x\, dx = 2\tan x - 3\sec x + C$

Second quotient. Same idea, with a $4$ in place of the $2$:

$\int \frac{4 - \sin x}{\cos^2 x}\, dx = 4\int \sec^2 x\, dx - \int \sec x\,\tan x\, dx = 4\tan x - \sec x + C$

Sum of sine and cosine, and basic antiderivatives

$\int \left(\sin x + \cos x\right)\, dx = -\cos x + \sin x + C = \sin x - \cos x + C$

Two quick antiderivatives:

$\int \cos 3x\, dx = \frac{\sin 3x}{3} + C, \qquad \int e^{2x}\, dx = \frac{e^{2x}}{2} + C$

Antiderivative with an initial condition

Given $f(x) = 4x^3 - 6$ with $f(0) = 3$, find the antiderivative $F(x)$:

$F(x) = \int \left(4x^3 - 6\right)\, dx = 4\cdot\frac{x^4}{4} - 6x + C = x^4 - 6x + C$

Apply $F(0) = 3$: substituting $x = 0$ gives $0 - 0 + C = 3$, so $C = 3$ and

$F(x) = x^4 - 6x + 3$

Key takeaways

• The antiderivative of $f'$ returns $f$, and a known value of the function fixes the constant of integration.
• Rewriting trigonometric ratios in terms of $\sec x$, $\tan x$ and the identity $\tan^2 x = \sec^2 x - 1$ turns awkward quotients into standard forms.
• $\int \sec^2 x\,dx = \tan x$ and $\int \sec x\,\tan x\,dx = \sec x$ are the two rules driving most of these integrals.