Integration as an Inverse Process of Differentiation

BASIC CONCEPT

Let F(x) be a differentiable function of x such that $\frac{d}{dx}\left[F\left(x\right)\right]=f\left(x\right)$. Then F(x) is called the integral of f(x). Symbolically, it is written as $\int$ f(x) dx = F(x).

f(x), the function to be integrated, is called the integrand.

F(x) is also called the anti-derivative (or primitive function) of f(x).

Constant of Integration:

As the differential coefficient of a constant is zero, we have

$\frac{d}{dx}\left(F\left(x\right)\right)=f\left(x\right)\Rightarrow \frac{d}{dx}\left[F\left(x\right)+c\right]=f\left(x\right)$.

Therefore, $\int f\left(x\right)dx$= F(x) + c.This constant c is called the constant of integration and can take any real value.

INTEGRATION AS THE INVERSE PROCESS OF DIFFERENTIATION

Basic formulae:

Anti-derivatives or integrals of some of the widely used functions (integrands) are given below.

$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n\Rightarrow \int x^{n}dx=\frac{x^{n+1}}{n+1}+c$, n $\ne$ – 1

$\frac{d}{dx}(\ln |x|)=\frac{1}{x}\Rightarrow \int \frac{1}{x}dx=\ln |x|+c$

$\frac{d}{dx}(e^x)=e^x\Rightarrow \int e^{x}dx=e^x+c$

$\frac{d}{dx}(a^x)=(a^x\ln a)\Rightarrow \int a^{x}dx=\frac{a^x}{\ln a}+c$ ( a > 0)

$\frac{d}{dx}(\sin x)=\cos x\Rightarrow \int \cos xdx=\sin x+c$

$\frac{d}{dx}(\cos x)=-\sin x\Rightarrow \int \sin xdx=-\cos x+c$

$\frac{d}{dx}(\tan x)={\sec }^{2}x\Rightarrow \int {\sec }^{2}xdx=\tan x+c$

$\frac{d}{dx}(\cos ecx)=(-\cot x\cos ecx)\Rightarrow \int \cos ecx\cot xdx=-\cos ecx+c$

$\frac{d}{dx}\left(\sec x\right)=\sec x\tan x\Rightarrow \int \sec X\tan x=\sec x+c...$

$\frac{d}{dx}(\cot x)=-\cos e{c}^{2}x\Rightarrow \int \cos e{c}^{2}xdx=-\cot x+c$

$\frac{d}{dx}({\sin }^{-1}\frac{x}{a})=\frac{1}{\sqrt{a^2-x^2}}\Rightarrow \int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$\frac{d}{dx}({\tan }^{-1}\frac{x}{a})=\frac{a}{x^2+a^2}\Rightarrow \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$

$\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}}\Rightarrow \int \frac{1}{\left|x\right|\sqrt{x^2-1}}dx={\sec }^{-1}(x)+c$

$\int \cot xdx=\int \frac{\cos x}{\sin x}dx=\ln |\sin x|+c$

$\int \tan xdx=-\int \frac{-\sin x}{\cos x}dx=-\ln |\cos x|+cor\ln |\sec x|+c$

$\int \sec xdx=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x}dx=\ln |\sec x+\tan x|+cor\ln \left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|+c$

$\int \cos ecxdx=\int \frac{\cos ecx(\cot x-\cos ecx)}{\cot x-\cos ecx}dx=\ln |(\cot x-\cos ecx)|+cor\ln \left|\tan \frac{x}{2}\right|+c$

Standard Formulae:

$\int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left|x+\sqrt{x^2+a^2}\right|+c$

$\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left|x+\sqrt{x^2-a^2}\right|+c$

$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+c$

$\int \sqrt{u^2+a^2}du=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln \left|u+\sqrt{u^2+a^2}\right|+c$

$\int \sqrt{u^2-a^2}du=\frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln \left|u+\sqrt{u^2-a^2}\right|+c$

$\int \sqrt{a^2-x^2}\text{ dx = }\frac{x}{\text{2}}\sqrt{a^2-x^2}+\frac{a^2}{2}\text{ si}{\text{n}}^{\text{-1}}\frac{x}{a}+c$

Illustration : Evaluate $\int \frac{\sin x}{1+\sin x}dx$.

Solution: $\frac{\sin x+1-1}{1+\sin x}=1-\frac{1}{1+\sin x}\frac{1-\sin x}{1-\sin x}$

= 1 –$\frac{1-\sin x}{1-{\sin }^{2}x}$ = 1 –$\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}$ = 1 – sec2 x – sec x tan x

Now $\frac{d}{dx}$(x – tan x – sec x) = 1 – sec2 x – sec x tan x

$\int \left(1-{\sec }^{2}x-\sec x\tan x\right)dx$ = x – tan x – sec x + c.