 Those are some special integrals whose results should not only come under substitution, they should also come under substitution. So, these are some special integrals whose results should not only come under substitution. These are some special integrals whose results should not only come under substitution. So, in cases where you have x square plus a square arising, the best substitution is put x as a tan, so it becomes integral of secant square theta d theta. When tan x is or tan theta is substituted as t or x is substituted as tan theta, d x will become a secant square theta d theta. In the denominator, you have a square tan square theta plus a square. So, if I take a square common in the denominator 1 plus tan square theta and c square and 1 plus tan square will get cancelled off. This will be left with 1 by a integral d theta, which is 1 by a theta plus c, which you can write it as 1 by a and theta from here will be tan inverse of x by a, but we want simple activity. Take 1 by a square, take 1 by a square common. So, you will get d x by 1 plus x by a whole square, correct? x by a divided by integral 1 by x square plus a square is given as 1 by tan inverse d x by theta. So, if I take 1 by a, so this is the second one. So, if I take 1 by a, then x minus a by a is equal to 0. So, if I take 1 by a, then x minus a by a is equal to 0. So, if I take 1 by a, then tan by 0 is equal to 0. So, if I take 1 by a plus, then what is the ideal substitution for such cases? A sin theta, A sin theta. So, we will again use x n A sin theta under the circumstances, d x becomes A sin theta tan theta d theta, correct? Right. Second, you substitute it as A sin theta. Now, if I substitute c theta tan theta d theta, yes or no, d theta, and then d theta, So, it becomes 1 by c theta by tan theta which becomes 1 by a cos theta d theta. What does it become cos theta d theta integration is ln ln mod cos theta cos theta minus mod theta. Now, I cannot leave my answer in terms of theta because theta is something which I have introduced in the problem by substituting x and a c theta. So, if I ask you if c theta is x by a 1 by h to the power of a, go c theta will be hypotenuse by yeah this will be what I am going to do. So, if I substitute a under root x square minus a square and what is cot theta? a by root over x square minus a square correct. Yes sir, so when I substitute it here minus a under root x square minus a square. Yeah. Now, if I write the x minus a term on top of a into root of x minus a and in the denominator I write it as root of x minus a into root of x plus a will that be fine. Yeah. So, these two terms will get cancelled. Half hour can be taken outside. So, you would realize that your expression would convert to 1 by 2 a ln x minus a by x plus a. And this is how we know dx by x square minus a square which is 1 by 2 a ln x minus a by x plus a. Is that okay? These are the things which you can do without using trigonometric substitution. Get from the problem and sit and derive these results. So, integration results can only be known through multiple practice. See, when the power is half you can write this by x plus a whole to the power of half right. This half I can put it down in front of the log of fractions that we can write this as x minus a into x plus a correct which can easily written as 1 by 2 a 1 by x minus a minus 1 by x plus a correct. So, instead of integrating this with respect to x we can choose to integrate this term with respect to a correct which clearly becomes ln mod x minus a minus ln mod x plus a which becomes 1 by 2 a ln mod x minus a by x plus a. Much easier way to solve this. Reach the same result. Now, see you don't have to. So, if I take a minus of this I will actually get my result. But normally the way it is written in the school textbooks is they take the minus as a power of this. So, it becomes ln x plus a by x minus a and in some times they will write a plus x by a minus x does not matter because your x minus a and a minus x within the mod sign will not matter by a minus x. If you don't understand this conversion you try writing this as a by x minus a plus b by x. So, d again I would substitute x as d theta in the denominator a get cancelled c theta integral is ln. Now, please note c theta would be x square by x square plus a square by a. So, your result would be ln x plus under root x square plus a square mod plus c this will be ln x. What happened? I got ln x by plus c that is nothing but ln x plus c. Try out e number question number e. What is the substitution? What will happen to d x? What will become a c theta d? That theta d x. And is this same thing minus a square d x. There is a lot of question is that there is a minus sign in the case of this. So, as in the problem even I substitute x as c theta a c theta. So, d x becomes a c theta tan theta d theta. So, denomination of this will become a c theta, but this time since your substitution is in terms of a c theta other time it was a tan theta. So, things will change slightly the c theta will become minus 1. Is it not? Which becomes ln of x plus under root x square minus a square plus c. Please note my log 6 1 answer. So, just write down these or if you have a formula list which you are maintaining add this formula to that. You should save initial few pages for writing the formula of this chapter. And this is x by a whole square. That is sin inverse. Which is 1 by a sin inverse x by a divided by a.