00:01
Okay, so we need to find the indefinite integrals.
00:04
The first integral is the integral of 1 over 4 plus x squared dx.
00:14
So if we rewrite this is the integral of 1 over 2 squared plus x squared dx, then it's in the form that we're used to for inverse tangent.
00:25
So this is 1 half, the inverse tangent of x over 2.
00:30
Do not forget your plus c or b we have the integral of one over the square root of 8 plus 2x minus x squared the x we can factor out that negative sign and then complete the square so this becomes the integral of 1 over the square root of negative x minus 1 squared minus 3 squared.
01:17
So that is adding and subtracting 1 squared to help us complete the square here.
01:27
And this is in the form of inverse sign.
01:31
So this is the inverse sign of x minus 1 over 3 plus c.
01:46
C, we have the integral of 1 over 1 plus sine of theta d theta, which is equal to the integral of 1 minus sine of theta over, so we're making the difference of two squares by multiplying and dividing by 1 minus sine of theta.
02:09
So 1 plus sine of theta, 1 minus sine of theta, d theta.
02:19
This is the integral of 1 minus sine of theta over 1 minus sine squared of theta, which d theta, which makes it the integral of 1 minus sine of theta over cosine squared of theta.
02:37
And i can split this up into two integrals.
02:40
The first integral is 1 over cosine squared of theta, which would be the integral of secan squared of theta, d theta, and then sine over cosine square...