Question

Some Special Integrals:
These are the some special Integrals which we can use for Integration..
Go through these formulas and must Remember them...

Answer 1

\[\int \frac{1}{x^2 + a^2}dx = \frac{1}{a}tan^{-1}\frac{x}{a} + C\]

\[\int \frac{1}{x^2-a^2}dx = \frac{1}{2a}Log\left | \frac{x-a}{x+a} \right | + C\]

\[\int \frac{1}{a^2-x^2}dx = \frac{1}{2a}Log\left | \frac{a+x}{a-x} \right | + C\]

\[\int \frac{1}{\sqrt{a^2-x^2}} = sin^{-1}\frac{x}{a} + C\]

\[\int \frac{1}{\sqrt{a^2+x^2}} = Log\left | x + \sqrt{a^2 + x^2} \right | + C\]

\[\int \frac{1}{\sqrt{x^2-a^2}} = Log\left | x + \sqrt{x^2 - a^2} \right | + C\]

Answer 2

thanks a lot

Answer 3

omg thank you XD

Answer 4

i cant find these in my books but problems that require them pop up in the questions

Answer 5

Here is one example too based on one of the formulas that I have written above:

Evaluate:

\[\color{green}{\int \frac{1}{\sqrt{9 - 25x^2}}dx}\]


Solution:

One must note that in the given formulas, the coefficient of x is 1 or -1..
Here, the coefficient of x is not 1 or -1, instead it is -25 or you can say 25..

So, firstly you should make it 1 by taking 25 common of the square root and one should also note that when a term is taken out of square root bracket, then it gets square rooted..

So 25 has the square root 5 so that there comes the factor 5..

Now, the integral becomes:

\[\huge \color{blue}{ = \int \frac{1}{5\sqrt{\frac{9}{25} - x^2}}dx}\]

\[\huge \color{red}{= \frac{1}{5} \int \frac{1}{\sqrt{(\frac{3}{5})^2 - x^2}}dx}\]

This resembles like the formula that I have written at number \(\color{violet}{\mathbf{4..}}\)

Using that Formula:

\[\huge \color{cyan}{= \frac{1}{5}(sin^{-1} \frac{x}{\frac{3}{5}} + C)}\]

\[\huge \color{orange}{= \frac{1}{5}sin^{-1} \frac{5x}{3} + \frac{C}{5}}\]

As \(\color{orange}{\frac{C}{5} = Constant = \textbf{We can take it as C itself..}}\)

So, the final answer becomes:

\[\huge \color{green}{= \frac{1}{5}sin^{-1} \frac{5x}{3} + C}\]

Therefore,

\[\huge \color{green}{{\int \frac{1}{\sqrt{9 - 25x^2}}dx} = \frac{1}{5}sin^{-1} \frac{5x}{3} + C}\]

Answer 6

However; you can derive them without memorising the above formula..

Answer 7

So in exams you will derive them..

Answer 8

naah. only when the table of integrals is not given :p

Answer 9

At every school or college, the Integral Table is not given,
so, it is better to remember these so that in emergency you can use them..

Answer 10

Thank you waterineyes!

Answer 11

If you can derive them it is very good..

Answer 12

well; the table of integrals is given in exams here. i just find deriving it is easier for harder problems..or when it is stated that you have to derive it..

Answer 13

Here??

Even I am no there..

You are thinking of only yours and not everyone..

Answer 14

@Mimi_x3 it is good if you think deriving them is easy and you can derive them...

Excellent..

Answer 15

Welcome @manita11 ..

Answer 16

It might be a good idea to copy the link in your profile so we can found this tutorial easier some time later @waterineyes, what do you think?

Answer 17

thanx water, very good post.. I derive them in exams because I'm too lazy to memorize haha

Answer 18

Yes I think the same...@knock .

Answer 19

@slaaibak ,
Bill Gates will give you the job because he likes Lazy person very much..
Ha ha ha..

Answer 20

can someone do a geometric proof?

Answer 21

do a proof of the standard integrals? how they got derived from?

Answer 22

yeah

Answer 23

well, im able to do it; it's just trig sub.

Answer 24

Well, I can prove tone or two of them, so that one can get better knowledge of these formulas where they have come from:

Let us prove First Formula..


\[\int\limits_{}^{}\frac{1}{x^2 + a^2}dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\]

Proof:

LHS:

Substitute \(\large x = atan \theta\) in Left Hand Side,

\(\large \theta = tan^{-1}\frac{x}{a}\)

\(\large dx = asec^2 \theta.d\theta\)

Putting in LHS,

\[\int\limits_{}^{}\frac{asec^2 \theta}{a^2 + a^2\tan^2 \theta}d \theta = \frac{1}{a}\int\limits_{}^{}\frac{\sec^2 \theta}{1 + \tan^2 \theta}d \theta\]

As, \(\large 1 + tan^2 \theta = sec^2 \theta\),

\[= \frac{1}{a}\int\limits_{}^{}1.d \theta = \frac{1}{a}\theta + C\]

Replace \(\theta\),

= \[\large \color{DarkGreen}{= \frac{1}{a}\tan^{-1}\frac{x}{a} + C}\]

\(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)

Answer 25

thats not a picture ; (

Answer 26

Let us prove now Last One...

\[\int\limits_{}^{}\frac{1}{\sqrt{x^2 - a^2}}dx = Log \left| x + \sqrt{x^2 - a^2} \right| + C\]

LHS:

Put \(\large x = asec \theta\),

\(\large \theta = sec^{-1} \frac{x}{a}\)

\(\large dx = sec \theta tan \theta.d \theta\)

Put in LHS:

\[\int\limits_{}^{}\frac{1}{a^2\sec^2 \theta - a^2}a.\sec \theta.\tan \theta.d \theta = \int\limits_{}^{}\sec \theta.d \theta = Log \left| \sec \theta + \tan \theta \right| + C_1\]

\[= Log \left| \sec \theta + \sqrt{\sec^2 \theta - 1} \right| + C_1 = Log \left| \frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1} \right| + C_1\]

[Since, \(\large 1 + tan^2 \theta = sec^2 \theta\)]

\[= Log \left| x + \sqrt{x^2 - a^2} \right| - Log(a) + C_1\]

\[\large \color{green} {= Log \left| x + \sqrt{x^2 - a^2} \right| + C}\]

\(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)