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`        Integrate the following and evaluate ∫2sinx/(3sinx+4cosx)`
one year ago

## Answers : (1)

```							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
```
one year ago
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