00:01
In this question, we are asked to calculate the given indefinite interval using trigonometric substitution.
00:06
And to do that, we first need to rewrite this interval a little bit.
00:10
So we want to have 1 in front of x squared.
00:17
To do that, we need to factor out 36.
00:19
So we are going to get the integral of the square root of 36 times x squared minus 49 over 36 over x -cube dx.
00:40
And this simplifies to the integral of 6 times the square root of x squared minus and we can rewrite 49 over 36 as 7 over 6 squared over x cubed d x now we are ready to make a trigonometric substitution and the substitution we are going to use here is x equals to 7 over 6 second theta and then d x equals to 7 over 6, second theta.
01:40
The derivative of second theta is second theta times tangent theta, right? d theta.
01:50
And now we can rewrite the integral as the integral of 6 times the square root of...
02:01
Now we will replace x by 7 over 6 second theta.
02:11
And it's squared minus 7 over 6 squared.
02:23
Divided by x cube right so here we will also replace x by 7 over 6 second theta we are going to get 7 over 6 second theta cube and we should not forget about d x d x is 7 over 6 second theta tangent theta d theta so i guess i need to move everything to the left a little bit otherwise i'm not going to fit and erase this so, d x is 7 over 6 second tangent d theta.
03:12
Second theta, tangent theta.
03:14
Well, unfortunately, i cannot fit d theta here.
03:21
What we can do next is cancel 6.
03:24
And we are going to get 7 times the integral of, now in the numerator, we can factor out 7 over 6 squared.
03:41
After factoring after factorization in the in parentheses we are going to get second squared theta minus one and this is the main reason we chose this substitution because second squared minus one is a trigonometric identity in the denominator it's going to be seven over six cube times second cube theta and time second theta times second theta times tangent theta d theta all right here we need to call that second squared theta minus 1 equals to tangent squared theta so we can replace this expression here by tangent squared theta and what we're going to get is 7 times the integral the square root of 7 over 6 squared becomes 7 over 6 times the square of and now we will replace second squared minus 1 by tangent squared in that and we can also cancel one power of second in the denominator it's still 7 over 6 cube by now 2nd squared theta and multiplied by tangent theta d theta.
05:32
We can cancel one power of 7 over 6 and we are going to get 7.
05:40
The reciprocal of 7 or 6 squared is 6 over 7 squared.
05:47
The square of tangent is tangent times tangent, divided by second squared d theta, now we need to simplify tangent squared over second squared and we can cancel also one power of seven right so we're going to get 36 over 7 times the integral of tangent squared over second squared and this equals to 36 over 7 times the integral now tangents are called a tangent squared is sine squared over cost squared and second squared is 1 over cost squared so this, and to divide two fractions, we need to multiply the fraction in the numerator by the reciprocal of the fraction in the denominator by cost squared over 1.
07:20
Cost squared cancels, and now we just need to calculate the interval of sine squared, which we can do by using double -angle formulas.
07:30
By the double -angle formulas, sine squared equals to one -half times, let's see, one minus cos 2 theta.
07:47
And this equals to 18 over 7 right because 36 over 2 equals to 18 times the integral of 1 minus cos 2 theta this equals to 18 over 7 the interval of 1 is theta and the integral of cost 2 theta is 1 half times sine 2 theta and d theta and finally this equals sorry there is no need in d theta anymore so plus the constant of so we are going to get 18 over 7 theta minus 9 over 7 and sine 2 theta is 2 sine theta times cos theta plus the constant of interracency.
09:02
Now this means that the original integral we are almost done...