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## Main Question or Discussion Point

int_0^(pi) sin(t)/[3 + cos(t)] dt

Answer:

u = cos(t); du = -sin(t) dt; a = 1; b = -1

int_1^(-1) -1/(u + 3) du

int_(-1)^1 1/(u + 3) du = ln(u + 3) from F(1) - F(-1)

ln(4) - ln(2) = ln(2)

I had help on this. And there are a few aspects I don't understand. What is the purpose of a=1; b=-1 and why does the integral separate into int_1^(-1) and then int_(-1)^1 why does it switch and what happened to pi and 0? and finally F(1)-F(-1). I know this is asking a lot but I am trying to grasp these concepts for the first time.

Thank you

Answer:

u = cos(t); du = -sin(t) dt; a = 1; b = -1

int_1^(-1) -1/(u + 3) du

int_(-1)^1 1/(u + 3) du = ln(u + 3) from F(1) - F(-1)

ln(4) - ln(2) = ln(2)

I had help on this. And there are a few aspects I don't understand. What is the purpose of a=1; b=-1 and why does the integral separate into int_1^(-1) and then int_(-1)^1 why does it switch and what happened to pi and 0? and finally F(1)-F(-1). I know this is asking a lot but I am trying to grasp these concepts for the first time.

Thank you