Indefinite Integral

Where there are no limits on the integral sign, the integral is called indefinite, meaning there is no specific value. Rather, the result is a family of functions. The integration is performed in the same way but we must remember to add an arbitrary constant known as the constant of integration. For example,

x2 dx = x3/3 + C

Why is this? If we take our answer x3/3 and differentiate with respect to x, we obtain x2. We would also obtain the same answer for x3/3 + 5 or x3/3 – 2 or x3/3 + any constant. Therefore, when we integrate, we have to add a constant because differentiation of a constant is zero.The value of the constant has to be determined by additional information about the equation, for example where it intercepts the y-axis.

Common Integrals

k f(u) du = kf(u)du

∫[f(ug(u)] du = ∫f(u) du ± ∫g(u) du

du=u + C

un du = un+1/(n+1) + C, (n ≠ -1)

du/u = ln|u| + C

eu du = eu + C

∫sin(u) du = – cos(u) + C

∫cos(u) du = sin(u) + C

∫tan(u) du = – ln|cos(u)| + C

∫cot(u) du = ln|sin(u)| + C

∫sec(u) du = ln|sec(u) + tan(u)| + C

∫cosec(u) du = ln|cosec(u) + cot(u)| + C

∫sec2(u) du = tan(u) + C

∫cosec2(u) du = -cot(u) + C

∫sec(u) tan(u) du =sec(u) + C

∫cosec(u) cot(u) du = – cosec(u) + C

∫(a2u2)-1/2du = – arcsin(u/a) + C

∫(a2 + u2)-1/2du = (1/a) arctan(u/a) + C

∫(a2u2)-1/2du = – arcsin(u/a) + C

d/dx[∫xaf(t)dt] = f(x)

Table 1. Common integrals.

Integration by Substitution

The substitution of a function, may simplify the integral allowing it to be calculated easily.


215x2 cos(x3) dx

Try u = x3, therefore, du = 3x2dx.

x2dx = (1/3) du

We must change the limits of integration, the new values come from u = x3, therefore when x= 1, u = 1 and when x= 2, u = 8. The integral becomes,

81(5/3) cos(u) du

(5/3) sin(u)|81 = (5/3)[sin(8) - sin(1)]

Integration by Parts

Let U and V be functions of x. From the product rule:

d(UV)/dx = V (dU/dx) + U (dV/dx)

Integrating both sides with respect to x and rearranging,

U(dV/dx).dx = UV – ∫ V (dU/dx) dx

Given some product to integrate, we arrange for U and dV to make the integral on the right-hand side, V(dU/dx) more simple than the integral U (dV/dx) we started with.


x exp(-x) dx

Let u =x, dv = exp(-x), therefore du = dx, v = -exp(-x)

x exp(-x) = -x exp(-x) + ∫-exp(-x) dx =

-x exp(-x) – exp(-x) + C

The Fundamental Theorem of Calculus

Integration is the inverse process of differentiation.

baf(x)dx = F(b)- F(a)

dF(x)/dx = f(x)

Therefore if we recognise that the function to be integrated as a derivative, then we can say the integral is the function that gaves that derivative. For this reason we sometime call the integral the anti-derivative.


When performing experimental work, we test the hypothesis and measure observable quantities. Often, we also plot graphs to demonstrate a relationship between the results and our theory.

A graph plots the relationship of one quantity against another on two axes at right-angles to each other. Usually, we have control over one of the quantities and this is known as the independent variable, the other quantity is determined by the outcome of the experiment or some mathematical relationship. We call this the dependent variable because it depends on the independent variable. We usually plot the independent variable on the x-axis and the dependent variable on the y-axis.

Наши рекомендации