00:02
6 .137, we're asked to derive the formula for the tangent to the int power, the integration of that.
00:10
So when i see this, i look at the integral as an n -minus 2.
00:13
So how could i take, and often the case with integration is, can i move things around in a way to manipulate to get something? so what if i were to write this as the integral of the tangent squared of x times a tangent of n -2 of x.
00:33
D x.
00:36
So if i do this then the integral stays the same.
00:39
But then i have a pythagorean identity.
00:41
I know that 1 plus the tangent squared of x is equal to the secant squared of x.
00:48
So i could rewrite all of this as the integral of, this would be what, sequent squared of x minus 1 times the tangent to the n minus 2x d x.
01:06
So i've got the integral tangent to the n of x d x .x.
01:13
Now this can be split into two different integrals, so this gives me the integral tangent to the nth power of x d x is equal to the integral secant squared of x times the tangent of n minus 2 of x d x minus the integral tangent n minus 2 of x d x now often the case with integration if you can get the integrand to show you a function and a derivative substitutions will typically work so what if z were equal to the tangent of x then d z would be the sequence squared of x d x d x so i can write this as the integral of, so this looks like z to the n minus 2 d z, and then minus the integral tangent n minus 2 of x d x...