00:01
In this question, we are asked to calculate the given integral.
00:03
To do that, we'll make the substitution x equals 6 tan theta.
00:11
Then dx equals 6 secant squared theta d theta.
00:18
Now we'll plug in this in the integral.
00:22
We'll get the integral of 6 cube multiplied by tan cube theta multiplied by the square root of 36 plus 36 tan squared theta multiplied by 6 secant squared theta d theta.
00:42
Then we can rewrite this as 6 to the fourth, the integral of 6 to the fourth multiplied by tan cube theta multiplied by the square root of 36 times 1 plus tan squared theta multiplied by secant squared theta d theta.
01:04
Then recall that 1 plus tan squared theta equals secant squared theta.
01:12
So we'll get the square root of 36 equals 6.
01:16
So we'll get 6 to the 5 multiplied by the integral of tan cube theta multiplied by secant theta because square root of secant squared equals secant multiplied by secant squared theta.
01:34
Next, we can rewrite this as 6 to the 5 times the integral of tan squared theta secant squared theta multiplied by tan theta secant theta d theta.
01:50
Then we'll make the substitution u equals secant theta.
01:57
And we'll also rewrite tan squared as secant squared minus 1.
02:05
Then du equals secant theta tan theta d theta.
02:15
This means we can replace this by du.
02:19
We can replace secant squared by u squared and tan squared by u squared minus 1 by using this formula.
02:28
So we'll get 6 to the 5 multiplied by the integral of u squared minus 1 multiplied by u squared du...