# How to find the integral

The concept of the integral is directly linked with the concept of primitive functions. In other words, to find the integral of this function, it is necessary to find such a function, with respect to which the original will be derived.

## Instruction how to find the integral

Step 1:

Integral refers to the concepts of mathematical analysis and graphically represents the area of the curvilinear trapezoid bounded by the x-axis limit points of integration. Finding integral functions is much more difficult than to find its derivative.

Step 2:

There are several methods for calculating indefinite integral: direct integration, the introduction of a differential sign, substitution method, integration by parts, substitution of Weierstrass, the Newton-Leibniz theorem, and others.

Step 3:

Direct integration involves bringing through simple transformations of the original integral to the table value. For example:∫dy / (sin²y·cos²y) = ∫(cos²y + sin²y) / (sin²y·cos²y) dy = ∫dy / sin²y + ∫dy / cos²y = -ctgy + tgy + C.

Step 4:

The method of introducing a differential mark or variable substitution is setting a new variable. At the same time the original integral is reduced to the new integral, which can be converted to a tabular view by direct integration: Let there be an integral ∫f (y) dy = F (y) + C and some variable v = g (y), then:∫f (y) dy -gt; ∫f (v) dv = F (v) + C.

Step 5:

It is necessary to remember some simple substitutions to facilitate the work with this method: dy = d (y + b); ydy = 1/2·d (y² + B); sinydy = - d (cosy); cosydy = d (siny).

Step 6:

Example:∫dy / (1 + 4·y²) = ∫dy / (1 + (2·y) ²) = [Dy -gt; d (2·y)] = 1/2·∫d (2·y) / (1 + (2·y) ²) = 1.2·arctg2·y + C.

Step 7:

Integration by parts produced according to the following formula:∫udv = u·v - ∫vdu.Primer:∫y·sinydy = [u = y; v = siny] = y·(-cosy) - ∫(-cosy) Dy = -y·cosy + siny + C.

Step 8:

The definite integral in most cases is located on the Newton-Leibniz theorem:∫f (y) dy in the interval [a; b] is equal to F (b) - F (a) Example: Find ∫y·sinydy the interval [0; 2π]:∫y·sinydy = [u = y; v = siny] = y·(-cosy) - ∫(-cosy) Dy = (-2π·cos2π + sin2π) - (-0·cos0 + sin0) = -2π.