The reason you substitute theta back is is because the original substitution mapped the function to a polar domain. The answer is correct for THAT domain, but it's not necessarily correct for the Euclidean domain the original problem was phrased in. So, what you do is remap the function back to a euclidean domain by substituting theta.

Is that what you were asking?

You use SineOppositeHypotenuse CosineAdjacentHypotenuse TangentOppositeAdjacent
If you subsituted sine in you need to solve for it in terms of x. Then you phrase it as a fraction (if its not a fraction just put a 1 under it). Then match sides using the trig functions. (Opposite top, Hypotenuse bottom for sine) draw a triangle and solve it in terms of x. You can then substitute that into any trig function to solve for it in terms of x. If you want an example I can work one out later.

Memorizing formula, of course!

integral(1/sqrt(a^2-x^2))=arcsin(x/a)

integral(1/sqrt(16-x^2))dx



using trig substitution

sin(-)=x/4
4sin(-)=x

sqrt(4^2-x^2)=4cos(-)

dx = d/dx(x)
dx = d/dx(4sin(-))
dx = 4cos(-)d(-)

substitute to original equation:

integral((1/4cos(-))(4cos(-)d(-)))
= integral (1 d(-))

integrate it and you get (-)+C.

now since sin(-)=x/4
(-) = arcsin(x/4)

therefore (-)+C = arcsin(x/4)+C

He used the triangle drawing to not have to explain what kind of triangle he was using. (as in, side length and stuff)