00:01
Let's now evaluate this given integral.
00:03
That is, we have to evaluate integral of sine inverse of x over 6x squared dx.
00:09
So in step one, we are going to replace the integration by parts method.
00:20
Integration by parts.
00:25
First, i'm going to rewrite this given integral as like that.
00:30
That is, sine inverse of x divided by 6 is square dx.
00:34
So this can be written as i write down this sign inverse of x as it is.
00:41
And we have divided by 6x squared.
00:45
So this could be written as 1 over 6x squared and then dx.
00:50
So basically we have written this as a product of two different functions.
00:56
The first function is an inverse trigonometric function.
01:00
And then the second function is an algebraic function.
01:02
So integration by parts method, if we have to follow this rule called i -l -a -t -e, where i stands for inverse trigonometric functions, l stands for logarithmic, a stands for algebra, and then t is for trigonometric and e for exponential.
01:24
So we have to see in this order that is from left to right.
01:28
So when we see left to right, the inverse function comes first.
01:32
So basically we follow this rule to choose the u.
01:36
And once we have chosen the u, rest of the parts will become the dv.
01:41
So when we follow this rule, since i comes per that is we have to look out for inverse trigonometric function.
01:48
Yes, we have a inverse trigonometric function that is sine inverse of x.
01:53
So which means we can choose u as.
01:57
So we choose u as the sine inverse of x.
02:01
And then rest of the terms, that is 1 over 6x squared dx, this is chosen as dv.
02:14
So i'm going to write dv.
02:17
This equals 1 over 6x squared dx.
02:23
Now let's consider this as separately.
02:26
Here we have to differentiate.
02:29
So we differentiate this side.
02:31
So differentiation of d, u is d .u, and differential of sine inverse of x is, when we use the standard formula, we will have 1 over under the root of 1 minus x squared and then we put the dx as well.
02:46
So basically we take differentials on both sides.
02:49
But on this side, we have to integrate so that we will get v.
02:54
So when we integrate, put this integral symbol, integral of dv is just v.
03:01
We have to integrate this 1 over 6x squared.
03:05
So i can factor this 1 over 6 and then relate this 1 over x squared as x raised to the power of negative 2 dx.
03:14
So this becomes when we apply the power rule of integral, this becomes x raised to the power of negative 2 plus 1 divided by the same number that is negative 2 plus 1.
03:25
And this term we will not have to put the constant of integration.
03:29
So just ignore that.
03:31
We will put that when we use the integration by parse formula.
03:35
So let's get this simplified.
03:36
So this is equal to 1 over 6 x raised to the power of negative 1 over negative 1.
03:42
So it's basically negative 1 over 6 times of x raise to the power of 1.
03:50
We can read it this as 1 over x.
03:52
So this is written simply as negative of 1 over 6x.
03:57
So this is v.
04:01
Now let's apply the integration by parts formula.
04:05
So i'm going to rewrite this original integral, that is sine inverse of x times of 1 over 6x squared dx.
04:17
As we considered this sine inverse of x as u and then the remaining term that is 1 or 6x squared as dv, we can apply the formula that is integral of u.
04:29
Dv and this equals uv minus integral of v d u so let me transfer this formula just above this expression so we will have it here so therefore when we apply this formula so this equals because this is in the form of u dv which means we apply this formula so according to the formula we have to multiply u and v we know that u is a sine inverse of x so i put sine inverse of x times of v.
05:05
We found the v as negative 1 over 6x square.
05:09
So i multiply by negative 1 over 6x squared minus integral of v that is negative 1 over 6x.
05:20
I'm sorry, this is 6x not 6x squared times of du.
05:27
So let's check du.
05:30
So this is the du.
05:31
That is 1 over under the root of 1 minus x squared d x so i write on this as 1 over 1 minus x squared this is under the root d x so we will simplify this so this is going to be we have a negative 1 here so i can rewrite this as negative of sine inverse of x in fact we can put a bracket over 6x and then we have a multiply we have to multiply the negatives that is we have negative times of negative is positive and i can factor this 1 over 6 and then put only the terms with x.
06:13
So therefore it is an integral of 1 over x times of under the root of 1 minus x squared t x.
06:22
Now we put this integration constant c.
06:27
So for the integral of this expression is negative sine inverse of x over 6x plus.
06:35
1 over 6 and then integral of 1 over x under the root of 1 minus x squared dx plus the integral constant c.
06:43
So this completes our step 1 and in step 2 we are going to focus on this integral alone.
06:50
That is we have to integrate this expression.
06:54
So let's proceed to step 2.
07:01
That is let's consider this integral separately.
07:05
That is we have to integrate 1 over x times of 1.
07:08
1 minus x squared under the root d x maybe we call this equation as one once we determine the integral of this one uh we come back here and replace this integral as we have found the result of this integral so let's call this integral as uh one i mean this we call this equation as equation one now let's focus on this and let's see how to determine this integral uh here i'm going to solve this integral by method of substitution.
07:40
That is, i'm going to put u equals under the root of 1 minus x squared.
07:46
Now i have to take the differentials on both sides.
07:49
In fact, i can consider this as 1 minus x squared.
07:53
I put this in exponent form as power 1 by 2.
07:56
Now if i take the differentials on both side, i will get the du equals.
08:00
I apply the power rule here.
08:02
That is, i put the power in front and then 1 minus x squared, raise to the power of 1 by 2 minus 1...