# Integral Formula

Integration is considered as the reverse process of differentiation, thus often integration is called as the inverse differentiation process. If a function f is differentiable in an interval in an interval I, i.e. derivative f exist at each point of I, then can we determine the function if the derivative f is known to us?

The answer to this question is yes! This function is known as the anti-derivative function, and the process of finding the anti-derivatives is known as the integration.

“Integral is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs.”

Integration is the basic operation of Integral Calculus. Integral formulas are classified based on,

• Integral formulas based on Rational functions
• Integral formulas based on Irrational functions
• Integral formulas based on Trigonometric functions
• Integral formulas based on Inverse trigonometric functions
• Integral formulas based on Hyperbolic functions
• Integral formulas based on Inverse hyperbolic functions
• Integral formulas based on Exponential functions
• Integral formulas based on Logarithmic functions
• Integral formulas based on Gaussian functions

### Below is the list of Integral Formulas based on the above functions,

$\large \int 1 \; dx = x + C$

$\large \int a \; dx = ax + C$

$\large \int x^{n} \; dx = \frac{x^{n+1}}{n+1} + C; \; n \neq -1$

$\large \int \sin x \; dx = – \cos x + C$

$\large \int \cos x \; dx = \sin x + C$

$\large \int \sec x = ln | \sec x + \tan x | + C$

$\large \int \sec x = ln | \sec x + \tan x | + C$

$\large \int \log x = x \log x – x + C$

$\large \int \sec ^{2} x \; dx = \tan x + C$

$\large \int \csc ^{2} x \; dx = -\cot x + C$

$\large \int \sec x (\tan x) \; dx = \sec x + C$

$\large \int \csc x (\cot x) \; dx = – \csc x + C$

$\large \int \frac{1}{x} \; dx = \ln |x| + C$

$\large \int e^{x} \; dx = e^{x} + C$

$\large \int a^{x} \; dx = \frac{a^{x}}{\ln a} + C; \; a> 0, a\neq 1$

$\large \int \frac{1}{\sqrt{1-x^{2}}} \; dx = \sin^{-1} x + C$

$\large \int \frac{1}{\sqrt{1+x^{2}}} \; dx = \tan^{-1} x + C$

$\large \int \frac{1}{|x|\sqrt{x^{2}-1}} \; dx = \sec^{-1} x + C$

$\large \int \sin^{n} (x) dx = \frac{-1}{n} \sin^{n-1} (x) \cos (x) +\frac{n-1}{n} \int \sin^{n-2} (x) dx$

$\large \int \cos^{n} (x) dx = \frac{1}{n} \cos^{n-1} (x) \sin (x) +\frac{n-1}{n} \int \cos^{n-2} (x) dx$

$\large \int \tan^{n} (x) dx = \frac{1}{n-1} \tan^{n-1} (x) +\int \tan^{n-2} (x) dx$

$\large \int \sec^{n} (x) dx = \frac{1}{n-1} \sec^{n-2} (x) \tan (x) + \frac{n-2}{n-1} \int \sec^{n-2} (x) dx$

$\large \int \csc^{n} (x) dx = \frac{-1}{n-1} \csc^{n-2} (x) \cot (x) + \frac{n-2}{n-1} \int \csc^{n-2} (x) dx$

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