Integrating exponential, trigonometric and rational functions

The standard integrals

• Every one of these is a derivative read backwards. Learn them as pairs.

$$\int e^x\,dx = e^x + c$$

$$\int \frac{1}{x}\,dx = \ln|x| + c$$

$$\int \cos x\,dx = \sin x + c$$

$$\int \sin x\,dx = -\cos x + c$$

$$\int \sec^2 x\,dx = \tan x + c$$

• The minus sign moves. Differentiating $\cos$ produces a minus; integrating $\sin$ produces one. Check which direction you are going before you write the sign.

The missing case from the power rule

• $\displaystyle\int x^n dx = \frac{x^{n+1}}{n+1}$ fails at $n = -1$ because the denominator would be zero.
• That gap is filled by the logarithm:

$$\int x^{-1}\,dx = \int \frac{1}{x}\,dx = \ln|x| + c$$

• The modulus signs matter. $\ln$ is only defined for positive numbers, but $\frac{1}{x}$ exists for negative $x$ too. The $|x|$ makes the result valid on both sides of the origin.

With a linear function inside

• Each standard integral has a version with $ax + b$ inside, and each one divides by $a$:

$$\int e^{ax+b}\,dx = \frac{1}{a}e^{ax+b} + c$$

$$\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b) + c$$

$$\int \sin(ax+b)\,dx = -\frac{1}{a}\cos(ax+b) + c$$

$$\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + c$$

$$\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \qquad (n \neq -1)$$

• Divide, do not multiply. Differentiation multiplies by $a$; integration undoes that by dividing. Getting it backwards is the single most common error on this page.

• Worked through:

  • $$\displaystyle\int e^{5x}dx = \frac{1}{5}e^{5x} + c$$
  • $$\displaystyle\int \sin 3x\,dx = -\frac{1}{3}\cos 3x + c$$
  • $$\displaystyle\int (2x-7)^4 dx = \frac{(2x-7)^5}{10} + c$$
  • $$\displaystyle\int \frac{1}{4x+1}dx = \frac{1}{4}\ln|4x+1| + c$$

• The report lists "successfully integrated a function of the type $(ax+b)^n$, i.e. could use the 'reverse chain rule'" at both Achieved and Merit level.

Rational functions

• A rational function is a polynomial divided by a polynomial. Three situations come up, and choosing the right one is the skill.

• Case 1 — single-term denominator: split it.

$$\int \frac{x^2 + 3}{x}\,dx = \int \left(x + \frac{3}{x}\right)dx = \frac{x^2}{2} + 3\ln|x| + c$$

• Case 2 — the numerator is the derivative of the denominator: use $\ln$.

$$\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$$

For example $\displaystyle\int \frac{2x}{x^2+1}dx = \ln\left|x^2+1\right| + c$, because $2x$ is exactly the derivative of $x^2+1$.

• Case 3 — the numerator is a constant multiple of the derivative: adjust.

$$\int \frac{x}{x^2+1}\,dx = \frac{1}{2}\int \frac{2x}{x^2+1}\,dx = \frac{1}{2}\ln\left|x^2+1\right| + c$$

• The critical warning. The report lists "integrated any function with $f(x)$ in the denominator to $\ln(f(x))$" among the Not Achieved behaviours. $\displaystyle\int \frac{1}{x^2+1}dx$ is not $\ln|x^2+1|$ — the numerator is 1, not $2x$. Always check that the top is (a multiple of) the derivative of the bottom before writing a logarithm.

Long division for improper fractions

• If the numerator's degree is greater than or equal to the denominator's, divide first:

$$\int \frac{x^2}{x+1}\,dx = \int \left(x - 1 + \frac{1}{x+1}\right)dx = \frac{x^2}{2} - x + \ln|x+1| + c$$

• The report lists "manipulate expressions including long division" among the Excellence behaviours.

Powers of $e$ with other things attached

• $\displaystyle\int e^{x^2}dx$ cannot be found with the methods in this standard — the multiplier $2x$ is missing.
• But $\displaystyle\int xe^{x^2}dx = \frac{1}{2}e^{x^2} + c$ can, because the $x$ supplies (half of) the needed multiplier. That is the reverse chain rule, covered next.