Integration

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Contents
  1. Definition
  2. Integration
  3. Polynomial
  4. Logarithm
  5. Exponential
  6. Sine
  7. Cosine
  8. Tangent
  9. INTEGRATION OF THE SQUARES OF THE CIRCULAR FUNCTIONS

Standard mathematical integrals

Calculus Maths

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Definition

In calculus an indefinite integral of a function f is a function F whose derivative is equal to $f$, i.e., $F' = f$. The process of solving for antiderivatives is called antidifferentiation (or indefinite integration) and its opposite function is called differentiation, which is the process of finding a derivative. Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.

For example

$I(x)=\int\;x\;dx=\frac{x^2}{2}+C$

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Integration

Polynomial

$\displaystyle \int x^n dx = \frac{x^{n\,+\,1}}{n\:+\:1}+C$ for all values of n except n = - 1

Example 1
Problem

$\int 4x^3+3x^2+1 dx$

Workings

Integral can written as : $4\int x^3dx+3\int x^2dx +x+C$

Solution

$x^4+x^3+x+C$

Logarithm

$\displaystyle \int\frac{1}{x}dx = \; Ln\:x + C$

Example 1
Problem

Find $I(x)=\int \frac{1}{x+a}dx\,\,\,a \in \mathhf{R}$

Workings

Therefore $\int \frac{1}{x+a} dx =Ln(x+a)+C$

Solution

$I=Ln(x+a)+C$

Exponential

$\displaystyle \int e^x\:dx = e^x + C$

Example 1
Problem

Find $I=\int 2e^{2x}dx$

Workings

We can see that $(e^{2x})'=2e^{2x}$

Solution

Therefore $I=e^{2x}+C$

Sine

$\displaystyle \int\sin\:x\:dx = \;-\:cos\:x+C$

Example 1
Problem

Find $I=\int \sin(2x)\;dx$

Workings

Therefore $I=-\frac{\cos(2x)}{2}+C$

Cosine

$\displaystyle \int cos\:x\:dx = \;sin\:x +C$

Example 1
Problem

Find $I=\int cos(2x)dx$

Workings

Therefore $I=\frac{sin(2x)}{2}+C$

Tangent

$\displaystyle \int tan\:x\:dx = \:-\:Ln\:cos\:x + C$

Example 1
Problem

Find $I=\int\;tan\;2x\;dx$

Workings

We can see that : $(-Ln\;cos(2x)+C)'=2tan(2x)$

Solution

Hence $I=\frac{-1}{2}Ln\;cos(2x)+C$

$\displaystyle \int sec^2\:x\:dx = tan\:x + C$

$\displaystyle \int\frac{1}{a^2\:+\:x^2}\:dx = \frac{1}{a}\:tan^{-1}\frac{x}{a} + C$

$\displaystyle \int\frac{1}{a^2\:-\:x^2}\:dx = \frac{1}{2a}\:Ln\:\frac{a\:+\:x}{a\:-\:x} = \frac{1}{a}\:tanh^{-1}\frac{x}{a}+C$

$\displaystyle \int\frac{1}{x^2\:-\:a^2}\:dx = \frac{1}{2\,a}\:Ln\,\left(\frac{x-a}{x+a} \right) = -\:\frac{1}{a}\:coth^{-1}\frac{x}{a}+C$

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$\displaystyle \int\frac{1}{\sqrt{(a^2\:-\:x^2)}}\:dx = sin^{-1}\:\frac{x}{a}+C$

$\displaystyle \int\frac{1}{\sqrt{(a^2+x^2)}}\:dx= Ln\left(x\:+\:\sqrt{(x^2+a^2)} \right) = sinh^{-1}\frac{x}{a}+C$

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$\displaystyle \int\frac{1}{\sqrt{(x^2\:-\:a^2)}}\:dx = Ln\left(x\:+\:\sqrt{(x^2\:-\:a^2)} \right) = cosh^{-1}\frac{x}{a}+C$

INTEGRATION OF THE SQUARES OF THE CIRCULAR FUNCTIONS

$\displaystyle \int sin^2(x)dx=\frac{1}{2}\:x\:-\:\frac{1}{4}\:sin\:2\,x + C$

$\displaystyle \int cos^2(x)dx=\frac{1}{2}\:x\:+\:\frac{1}{4}\:sin\:2\,x + C$

$\displaystyle \int tan^2\:x\:dx = (tan\:x)\:-\:x + C$

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$\displaystyle \int cot^2\:x\:dx = -\:(cot\:x)\:-\,x + C$

$\displaystyle \int cosec^2\:x\:dx = -\:cot\:x + C$