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There are two sides to the mighty coin of calculus: Derivatives and Integrals. There is a whole study set for the former one, but this one goes to the latter one. The idea of an integral has been made pretty clear and explicit by now.
But we cannot forget formulas, the very basis of mathematics. Integration is littered with a multitude of distinct formulas, which makes it an easy task to do an integral, partly.
Let's take a look at some of these Integration Formulas:
It might sound absurd at first, that we can have formulas for integrals straight away. But there is a catch, we can only have formulas for relatively simple integrals, not for more composite functions.
Let us start with the very basics, the fundamental formulas of integration:
The following formulas we are going to talk about, are the very fundamental formulas, upon which other formulas are eventually built.
Hence, it would be a good practice to remember them by heart:
$$ \begin{aligned} &\int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1}+c \\ &\int e^x \, \mathrm{d}x=e^x+c \\ &\int a^x \, \mathrm {d}x=\frac{a^x}{\log_e a}+c \\&\int \frac{1}{x} \, \mathrm{d}x=\log_e x +c \end{aligned}$$
We just saw integration formulas for functions such as the linear function, quadratic functions, exponential function, logarithmic functions, and also functions like \(\frac{1}{1+x}, \ \frac{1}{1+x^2}\) and so on.
But there is a whole class of important functions that we have not addressed yet, the Trigonometric Functions. There are some basic formulas one needs to remember for trigonometric functions, from which we can solve more complicated trigonometric functions. These Integration formulas are as follows:
$$ \begin{aligned} &\int \cos x \, \mathrm{d}x=\sin x+c \\ &\int \sin x \, \mathrm{d}x=-\cos x+c \\ &\int \tan x \, \mathrm{d}x=-\log_e |\cos x|+c \\ &\int \cot x \, \mathrm{d}x=\log_e |\sin x|+c \\ &\int \sec^2 x \, \mathrm{d}x=\tan x +c \\ &\int \csc x \, \mathrm{d}x=\log_e |\frac{\tan x}{2}|+c \\ &\int \sec x \tan x \, \mathrm{d}x=\sec x+c \\ &\int \csc^2 x \, \mathrm{d}x=-\cot x+c \\ &\int \csc x \cot x \, \mathrm{d}x=-\cot x+c \end{aligned}$$
You might wonder why there is a constant of integration at the end of every indefinite integral. The reason is that if we take the derivative of the result we get, the constant disappears, this is because the derivative is the opposite of an integral.
As you may have noticed, the integration we saw are all related to indefinite integrals. But what about definite integrals? The integrals of all the functions remain unchanged, the only thing introduced are the limits of integration.
Below are some formulas, properties essentially, that are crucial while doing definite integration.
$$ \begin{aligned} &\int_a^b f(x) \, \mathrm{d}x=F(b)-F(a) \ ; \text{where} \ \int f(x) \, \mathrm{d}x=F(x)+c \\ &\int_a^b f(x) \, \mathrm{d}x=-\int_b^a f(x) \, \mathrm{d}x \\ &\int_a^b f(x) \, \mathrm{d}x=\int_a^b f(a+b-x) \, \mathrm{d}x \\ &\int_{-a}^a f(x) \, \mathrm{d}x=2\int_0^a f(x)\, \mathrm{d}x\ ; \text{where} \ f(x) \ \text{is an even function} \\ &\int_{-a}^a f(x) \, \mathrm{d}x=0 \ ; \text{where} \ f(x) \ \text{is an odd function} \\ \end{aligned}$$
Notice that there is no constant of integration in definite integration.
While solving integrals, we need to know a multitude of different integration formulas, without which, solving integrals will be impossible (unless you cheat using WolframAlpha or something!).
Below is a list of almost all the integration formulas one shall need in order to solve more complicated integrals, which comprise different composite functions. This will include formulas you already saw earlier (trigonometric, fundamental) alongside inverse trigonometric formulas, hyperbolic formulas, and many more. It is a handy chart one should have while doing integrals:
Let us revisit some of the fundamental integration formulas as a chart:
$$ \begin{aligned} &\int x^n\, \mathrm{d} x=\frac{x^{n+1}}{n+1}+c \\ &\int e^x \, \mathrm{d}x=e^x+c \\ &\int a^x \, \mathrm{d}x=\frac{a^x}{\log_e a}+c \\ &\int \frac{1}{x} \, \mathrm{d}x=\log_e x +c \\ &\int k\, \mathrm{d} x=k x+c \end{aligned}$$
Below is the chart for integrals of inverse trigonometric functions:
$$ \begin{aligned} &\int \sin^{-1} x \, \mathrm{d} x=x \sin^{-1} x+\sqrt{1-x^2}+c \\&\int \cos^{-1} x \, \mathrm{d} x=x \cos^{-1} x-\sqrt{1-x^2}+c \\&\int \tan^{-1} x \, \mathrm{d} x=x \tan^{-1} x-\frac{1}{2} \ln \left(x^2+1\right) +c\\&\int \cot^{-1} x \, \mathrm{d} x=x \cot^{-1} x+\frac{1}{2} \ln \left(x^2+1\right) +c\\ &\int \csc ^{-1} x \, \mathrm{d} x=x \csc ^{-1} x+\ln \left|x+\sqrt{x^2-1}\right|+c \\ &\int \sec ^{-1} x \, \mathrm{d} x=x \sec ^{-1} x-\ln \left|x+\sqrt{x^2-1}\right|+c \end{aligned}$$
There are some other integration formulas, which are listed below: $$ \begin{aligned} &\int e^{c x} \sin b x \, \mathrm{d} x=\frac{e^{c x}}{c^2+b^2}(c \sin b x-b \cos b x) +c\\ &\int e^{c x} \cos b x \, \mathrm{d} x=\frac{e^{c x}}{c^2+b^2}(c \cos b x+b \sin b x) +c\\&\int \frac{1}{x^2+a^2} \, \mathrm{d} x=\frac{1}{a} \tan^{-1} \frac{x}{a} +c\\&\int \frac{1}{x^2-a^2} \, \mathrm{d} x= \left\{\begin{array}{l} \displaystyle \frac{1}{2 a} \ln \left( \frac{a-x}{a+x} \right) +c\\\displaystyle \frac{1}{2 a} \ln \left( \frac{x-a}{x+a} \right)+c \end{array}\right. \end{aligned}$$
Let us revisit the trigonometric formulas with some squared formulas and hyperbolic formulas as well:
$$ \begin{aligned} &\int \sin x \, \mathrm{d} x=-\cos x+c \\&\int \cos x \, \mathrm{d} x=\sin x+c \\&\int \tan x \, \mathrm{d} x=\ln |\sec x|+c \\&\int \sec x \, \mathrm{d} x=\ln |\tan x+\sec x|+c \\&\int \sin ^2 x \, \mathrm{d} x=\frac{1}{2}(x-\sin x \cos x)+c \\&\int \cos ^2 x \, \mathrm{d} x=\frac{1}{2}(x+\sin x \cos x)+c \\&\int \tan ^2 x \, \mathrm{d} x=\tan x-x+c \\&\int \sec ^2 x \, \mathrm{d} x=\tan x+c \\ &\int \sin x \cos ^n x \, \mathrm{d} x=-\frac{\cos ^{n+1} x}{n+1}+c \\ &\int \sin ^n x \cos x \, \mathrm{d} x=\frac{\sin ^{n+1} x}{n+1} +c\\&\int \sinh x \, \mathrm{d}x=\cosh x +c \\ &\int \cosh x \, \mathrm{d}x=\sinh x +c\end{aligned} $$
Some of the above formulas are themselves derived from the other ones, we take them as formulas since they appear quite often while solving integrals.
Now let us apply these formulas and solve a couple of integrals. We shall use the above formulas without providing their proof, unless we are asked to provide one specifically.
Solve the indefinite integral \( \displaystyle \int \cos 4x \, \mathrm{d}x\).
Solution:
Recall that \(\int \cos x \, \mathrm{d}x=\sin x+c\), we get here:
$$\int \cos 4x \, \mathrm{d}x= \frac{\sin 4x}{4}+c \ \ \ \ \ (\because \int f(ax)\, \mathrm{d}x=\frac{F(ax)}{a}+c)$$
Evaluate the definite integral \( \displaystyle \int_0^1 (x+1)^2 \, \mathrm{d}x\).
Solution:
Here first we will integrate it as we normally would, and then apply the limits of integration using the property of definite integration (\(\int_a^b f(x) \, \mathrm{d}x=F(b)-F(a)\)).
Using the formula \( \displaystyle \int x^n \, \mathrm{d}x=\frac{x^{n+1}}{n+1}+c\),
$$ \begin{aligned} \int_0^1 (x+1)^2 \, \mathrm{d}x &=\left[\frac{(x+1)^3}{3} \right]_0^1 \\ &=\frac{(1+1)^3}{3}-\frac{(0+1)^3}{3} \\ &=\frac{8-1}{3} \\ \therefore \int_0^1 (x+1)^2 \, \mathrm{d}x &=\frac{7}{3} \end{aligned}$$
Evaluate the integral \(\int \cos^2 x \, \mathrm{d}x\).
Solution:
Firstly, we shall use the double-angle trigonometric identity \(\cos^2 x=\frac{1+\cos 2x}{2}\):
$$\int \cos^2 x \, \mathrm{d}x=\int \frac{1+\cos 2x}{2} \, \mathrm{d}x$$
Now, using the integral formulas \(\int x^n \, \mathrm{d}x=\frac{x^{n+1}}{n+1}+c\) and \(\int \cos x \, \mathrm{d}x=\sin x+c\):
$$\begin{aligned} \int \cos^2 x \, \mathrm{d}x &=\int \frac{1+\cos 2x}{2} \, \mathrm{d}x \\ &=\frac{x}{2}+\frac{\sin 2x}{4}+c \end{aligned}$$
Evaluate the integral \(\int_{-\pi}^{\pi} \sec x \tan x \, \mathrm{d}x\).
Solution:
Notice that the limits of integration are of the form \(-a\) to \(a\), which means we should first check whether the given function is even or odd.
$$ \begin{aligned} f(-x) &=\sec (-x) \tan (-x) \\ &=-\sec x \tan x \\ \therefore f(-x) &=-f(x) \end{aligned}$$
Hence, the function is an odd function. So we shall use the property of the definite integral: \(\int_{-a}^a f(x) \, \mathrm{d}x = 0\).
Thus,
$$\int_{-\pi}^{\pi} \sec x \tan x \, \mathrm{d}x =0$$
Evaluate \(\int_{\pi / 3}^{\pi / 2} x \sin x\, \mathrm{d} x\)
Solution:
Notice that we have the integrand as a product of two functions, so we have to use integration by parts:
\( \begin{align} &u=x \ \, \, \text{and} \, \, \mathrm{d} v=\sin x \, \mathrm{d}x \\ &\, \mathrm{d}u= \mathrm{d} x \, \, \ \text{and} \quad v=-\cos x \end{align}\)
which yields
$$ \begin{aligned} \int x \sin x \, \mathrm{d}x &=-x \cos x-\int-\cos x \, \mathrm{d}x \\ &=-x \cos x+\sin x+c \end{aligned} $$
Hence,
$$ \begin{aligned} \int_{\pi / 3}^{\pi / 2} x \sin x \, \mathrm{d}x &=[-x \cos x+\sin x]_{\pi / 3}^{\pi / 2} \\ &=\left[\left(-\frac{\pi}{2}\right)\left(\cos \frac{\pi}{2}\right)+\sin \frac{\pi}{2}\right]-\left[\left(-\frac{\pi}{3}\right)\left(\cos \frac{\pi}{3}\right)+\sin \frac{\pi}{3}\right] \\ &=(0+1)-\left(-\frac{\pi}{6}+\frac{\sqrt{3}}{2}\right) \\ &=1+\frac{\pi}{6}-\frac{\sqrt{3}}{2} \\ &=\frac{6-3 \sqrt{3}+\pi}{6} \end{aligned}$$
The 3 basic integration methods are Integration by Substitution, Integration by Parts and Direct Integration.
Basic integration involves the three basic integration techniques alongside the basic integration formulas.
The integral of dx is x+c where c is a constant.
Integration is the opposite operation of taking a derivative, the derivative of a constant disappears, so we need to add it back when we integrate the function.
The rules of integration are quite basic, following the integration formulas and using the three basic integration methods, and applying the laws of definite integration when the limits are given.
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