anonymous
  • anonymous
Some Special Integrals: These are the some special Integrals which we can use for Integration.. Go through these formulas and must Remember them...
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
katieb
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anonymous
  • anonymous
\[\int \frac{1}{x^2 + a^2}dx = \frac{1}{a}tan^{-1}\frac{x}{a} + C\] \[\int \frac{1}{x^2-a^2}dx = \frac{1}{2a}Log\left | \frac{x-a}{x+a} \right | + C\] \[\int \frac{1}{a^2-x^2}dx = \frac{1}{2a}Log\left | \frac{a+x}{a-x} \right | + C\] \[\int \frac{1}{\sqrt{a^2-x^2}} = sin^{-1}\frac{x}{a} + C\] \[\int \frac{1}{\sqrt{a^2+x^2}} = Log\left | x + \sqrt{a^2 + x^2} \right | + C\] \[\int \frac{1}{\sqrt{x^2-a^2}} = Log\left | x + \sqrt{x^2 - a^2} \right | + C\]
mathslover
  • mathslover
thanks a lot
alexwee123
  • alexwee123
omg thank you XD

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alexwee123
  • alexwee123
i cant find these in my books but problems that require them pop up in the questions
anonymous
  • anonymous
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, \[\huge \color{green}{{\int \frac{1}{\sqrt{9 - 25x^2}}dx} = \frac{1}{5}sin^{-1} \frac{5x}{3} + C}\]
Mimi_x3
  • Mimi_x3
However; you can derive them without memorising the above formula..
anonymous
  • anonymous
So in exams you will derive them..
Mimi_x3
  • Mimi_x3
naah. only when the table of integrals is not given :p
anonymous
  • anonymous
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..
anonymous
  • anonymous
Thank you waterineyes!
anonymous
  • anonymous
If you can derive them it is very good..
Mimi_x3
  • Mimi_x3
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..
anonymous
  • anonymous
Here?? Even I am no there.. You are thinking of only yours and not everyone..
anonymous
  • anonymous
@Mimi_x3 it is good if you think deriving them is easy and you can derive them... Excellent..
anonymous
  • anonymous
Welcome @manita11 ..
anonymous
  • anonymous
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?
slaaibak
  • slaaibak
thanx water, very good post.. I derive them in exams because I'm too lazy to memorize haha
anonymous
  • anonymous
Yes I think the same...@knock .
anonymous
  • anonymous
@slaaibak , Bill Gates will give you the job because he likes Lazy person very much.. Ha ha ha..
UnkleRhaukus
  • UnkleRhaukus
can someone do a geometric proof?
Mimi_x3
  • Mimi_x3
do a proof of the standard integrals? how they got derived from?
UnkleRhaukus
  • UnkleRhaukus
yeah
Mimi_x3
  • Mimi_x3
well, im able to do it; it's just trig sub.
anonymous
  • anonymous
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 \(\theta\), = \[\large \color{DarkGreen}{= \frac{1}{a}\tan^{-1}\frac{x}{a} + C}\] \(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)
UnkleRhaukus
  • UnkleRhaukus
thats not a picture ; (
anonymous
  • anonymous
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, \(\large 1 + tan^2 \theta = sec^2 \theta\)] \[= Log \left| x + \sqrt{x^2 - a^2} \right| - Log(a) + C_1\] \[\large \color{green} {= Log \left| x + \sqrt{x^2 - a^2} \right| + C}\] \(\large \color{Orange}{\mathbf{Hence \; Proved..}}\)
UnkleRhaukus
  • UnkleRhaukus
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