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

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

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