# Integrate the following and evaluate ∫ 2sinx /(3sinx+4cosx)

Arun
25750 Points
6 years ago
You want to split it up in a very specific way. On the one hand, we can integrate the function:

(3cos(x) + 4sin(x)) / (3cos(x) + 4sin(x)) = 1

really easily (it just integrates to x). On the other hand, we can integrate the function:

(-3sin(x) + 4cos(x)) / (3cos(x) + 4sin(x))

easily as well, because a substitution of u = 3cos(x) + 4sin(x) yields an integral of ln|3cos(x) + 4sin(x)|. So, if we can express the numerator 2sin(x) + 3cos(x) as a linear combination of 3cos(x) + 4sin(x) and -3sin(x) + 4cos(x), then we can split the fraction up, and integrate each part easily.

Suppose A and B are such that:

A(3cos(x) + 4sin(x)) + B(-3sin(x) + 4cos(x)) = 2sin(x) + 3cos(x)

for all x. Then, equating coefficients:

4A - 3B = 2
3A + 4B = 3

Solving simultaneously yields:

A = 17/25
B = 6/25

Therefore the function is equal to:

17/25 + (6/25) * (-3sin(x) + 4cos(x)) / (3cos(x) + 4sin(x))

which integrates to:

(17x + 6ln|3cos(x) + 4sin(x)|) / 25 + C