 integral 6x to 15x square minus 7x plus 6 by 2x square minus 2x plus 1, then this will be 1 plus 2b to x into this and then there is just one part of the order, kept this constant of the quadratic, the constant of the quadratic right, the constant of the quadratic is 3, this is rdg, this is rdg, rdg, that is object demand right, v3 is 1, v2 is 1, v2 is the term with the x square no, that is correct, v3 is 2, v3 is 2, okay let me do this, differentiate both sides and respect to x, so left side will become, 4x minus 2 and the 2 below will look like we will get cancel, this 2 I can, any other term, from here I 20 is equal to 4b1, x term minus 7 on the left, right side I will get minus 2b1, so let us put the value, so minus rdg v3, v3, v3 is a constant, but at x is 6, the constant here is b1, the constant here will be coefficient of x, so this will be 2 root 2, the answer x term minus x plus half, so your final answer is 3 plus 5x under root 2x square minus 2x plus 1, so under root the initial function, under root the initial function, under root the initial function, that is what we learned in the standard integrals, type called ph5, those are names, 1 by a plus b cos x whole square, right and 1 by a plus b sin x whole square and we could not take any problem on that, so we will take some problems on that, I took the root for problem, okay so we will take up a problem, then not always, when there is a sin x here then cos x, if there is a cos x, if this is sin you take cos by this term itself, without the square, if this is a cos you take sin divided by, then you do dp by dx, when you do dp by dx here you would apply Cauchy rule, 16 sin x minus 9, correct, then just for the purpose of simplification, y is equal to 16 plus 9 sin x, here it says you can make it, but just for the purpose of simplification you can take minus, no you cannot cancel, minus and cancel, okay, so here sin x can be taken as y minus 16 by 9, okay, so just put it into this expression, just for the purpose of simplification, y doesn't, it is just helping you to simplify the expression, so minus 16 in place of sin x you can write minus 16 by 9, minus 9 by y square, minus 16 by 9 y to 56 by 9 minus 9 by y square, so 56 minus 81, how much is it, 1, 170, so minus 175 by 9 y square, so what should be the class? I have just taken shortcuts, you open it in terms of and this you would just keep it as 175 by 9 i, so that later on we can make i the subject of the formulas, right, so p is your formula, okay, so you can write minus 16 by 9 y square, minus 16 by 9 y square, okay, so correct, so I get this term by writing sin x as tan x by 2 by 1 plus tan x by 2, so take sin x as square x by 2, so 1 plus tan x square x by 2 we go on the numerator which will become c square x by 2, is that it, now what is the next stage, take tan x by 2 as 80 plus 1, correct, yes or no, so it is 1 by 4 square minus 66 plus 1, 56 minus 81, 175, so root 1 by 16 by 16 by 16 by 16, so x by will be 16 t plus 9 by root 170, so this will go by a factor of, so we will have 16 plus 9 sin x minus 16 by 9 times root 175, tan inverse 16 t, what is t a, tan x by 2 plus 9 by root 175 plus 170, so your i will be 9 by 175 times cos x plus 16 plus 9 sin x plus 32 by 9, okay.