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

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


1 Answers

25763 Points
3 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

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