Integration is a fundamental calculus concept where one has to find a function corresponding to its derivative. It is a method of combining parts to find the whole. Integration formulas can be used to solve different types of integration problems in algebra, trigonometry, exponential functions, logarithms, and inverse trigonometry. They consist of trigonometric ratios, product of functions, inverse trigonometric functions and advanced integration formulas. It is a reverse process of differentiation. So, if d/dx (y) = z, then ∫zdx = y. The fundamental integration formula, which serves as the basis for serving all integration formulas is
∫ f'(x)dx = f(x) + C
Integral Formulas are classified based on the following functions:
The integration by parts formula is applied when the given function can be described as the product of two functions:
∫ f(x)g(x) dx = f(x)∫ g(x) dx - ∫ (∫f'(x)g(x) dx) dx + C
Integration by substitution is applied when a function is a function of another function. That is, if I = ∫ f(x) dx, where x = g(t) such that dx/dt = g'(t), then dx = g'(t)dt:
I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt
The formula for Integration by Partial Fractions is used when the integral of P(x)/Q(x) is required and P(x)/Q(x) is an improper fraction:
P(x)/Q(x) = R(x) + P1(x)/ Q(x)
where:
The formula for Definite integral is used when the integration limit is given. In definite integration, the solution to the question is a constant value. Definite integration is solved as:
∫ab f(x) dx = F(b) - F(a)
The formula for Indefinite Integration Formula is used to solve the indefinite integration when the limit of integration is not given. In indefinite integration, we use the constant of the integration which is denoted by C:
∫ f(x) dx = F(x) + C
∫ cosx dx = sinx + C
∫ sinx dx = -cosx + C
∫ sec2x dx = tanx + C
∫ csc2x dx = -cotx + C
∫ secxtanx dx = secx + C
∫ cscxcotx dx = -cscx + C
∫ tanx dx = ln|secx| + C
∫ cotx dx = ln|sinx| + C
∫ secx dx = ln|secx + tanx| + C
∫ cscx dx = ln|cscx - cotx| + C
∫ 1/(1 - x^2) dx = sin-1x + C
∫ -1/(1 - x^2) dx = cos-1x + C
∫ 1/(1 + x^2) dx = tan-1x + C
∫ -1/(1 + x^2) dx = cot-1x + C
∫ 1/(x√(x^2 - 1)) dx = sec-1x + C
∫ -1/(x√(x^2 - 1)) dx = csc-1x + C
Given: ∫ 1/(x^2 + a^2) dx
Formula used: ∫ 1/(x^2 + a^2) dx = (1/a) tan-1(x/a) + C
Solution: Applying the formula: ∫ 1/(x^2 + a^2) dx = (1/a) tan-1(x/a) + C
Given: ∫ √(x^2 - a^2) dx
Formula used: ∫ √(x^2 - a^2) dx = (x/2) √(x^2 - a^2) - (a^2/2) log|x + √(x^2 - a^2)| + C
Solution: Applying the formula: ∫ √(x^2 - a^2) dx = (x/2) √(x^2 - a^2) - (a^2/2) log|x + √(x^2 - a^2)| + C
Given: ∫ 1/√(1 - x^2) dx
Formula used: ∫ 1/√(1 - x^2) dx = sin-1x + C
Solution: Applying the formula: ∫ 1/√(1 - x^2) dx = sin-1x + C
These examples illustrate the application of integration formulas to find antiderivatives of various functions. Each formula is specialized to handle different types of integrals, facilitating efficient problem-solving in calculus and beyond.
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