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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{1}{x^2 + a^2}dx = \frac{1}{a}tan^{1}\frac{x}{a} + C\] \[\int \frac{1}{x^2a^2}dx = \frac{1}{2a}Log\left  \frac{xa}{x+a} \right  + C\] \[\int \frac{1}{a^2x^2}dx = \frac{1}{2a}Log\left  \frac{a+x}{ax} \right  + C\] \[\int \frac{1}{\sqrt{a^2x^2}} = sin^{1}\frac{x}{a} + C\] \[\int \frac{1}{\sqrt{a^2+x^2}} = Log\left  x + \sqrt{a^2 + x^2} \right  + C\] \[\int \frac{1}{\sqrt{x^2a^2}} = Log\left  x + \sqrt{x^2  a^2} \right  + C\]

alexwee123
 4 years ago
Best ResponseYou've already chosen the best response.0i cant find these in my books but problems that require them pop up in the questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here 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, \[\huge \color{green}{{\int \frac{1}{\sqrt{9  25x^2}}dx} = \frac{1}{5}sin^{1} \frac{5x}{3} + C}\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0However; you can derive them without memorising the above formula..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So in exams you will derive them..

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0naah. only when the table of integrals is not given :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0At 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you waterineyes!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you can derive them it is very good..

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0well; 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here?? Even I am no there.. You are thinking of only yours and not everyone..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Mimi_x3 it is good if you think deriving them is easy and you can derive them... Excellent..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It 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?

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.0thanx water, very good post.. I derive them in exams because I'm too lazy to memorize haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes I think the same...@knock .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@slaaibak , Bill Gates will give you the job because he likes Lazy person very much.. Ha ha ha..

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0can someone do a geometric proof?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0do a proof of the standard integrals? how they got derived from?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0well, im able to do it; it's just trig sub.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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 \(\theta\), = \[\large \color{DarkGreen}{= \frac{1}{a}\tan^{1}\frac{x}{a} + C}\] \(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0thats not a picture ; (

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let 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, \(\large 1 + tan^2 \theta = sec^2 \theta\)] \[= Log \left x + \sqrt{x^2  a^2} \right  Log(a) + C_1\] \[\large \color{green} {= Log \left x + \sqrt{x^2  a^2} \right + C}\] \(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341765523705:dw
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