00:01
So we wanted to get this in the form of a squared minus u squared to u because only this and various variations of this can we actually integrate using a special formula.
00:12
And looking at this, so 16 is actually a perfect square, meaning we can assume that a equals 4, right? similarly with 4x squared, right? therefore, u equals 2x.
00:25
However, we also need to get to you, right? so if you take that derivative of both sides, we find that to you, equals 2 dx, right? or also, or dx equals du over 2, right? so once we have these values, we can now substitute the values of x, the values of u and a for the values of x right here, right? so we can get that in this form, right? and once we get that, we end up with one -half times the integral of group a squared minus u squared due, right? so once we get that, we can now apply this to the formula that we know very well, right? which is one -half times u -route a -squared minus u -squared plus a squared arc -sign of u over a, right, plus c, right? so we got a equals 4 and u equals 2x, right? and then once we plug these values into the formula right here, we can't forget the one half earlier from our integration right here, right? and then once we plug everything in, we get one fourth times 2x times root 16 minus 4x squared plus 16 arc sine of 2x...