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Please evaluate the following integral with steps, if possible: 1/ [3 sin(x) - 4 cos(x)]dx.

Thank you very much.

Guest Jun 16, 2017
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Take the integral:
 integral1/(3 sin(x) - 4 cos(x)) dx
For the integrand 1/(3 sin(x) - 4 cos(x)), substitute u = tan(x/2) and du = 1/2 dx sec^2(x/2). Then transform the integrand using the substitutions sin(x) = (2 u)/(u^2 + 1), cos(x) = (1 - u^2)/(u^2 + 1) and dx = (2 du)/(u^2 + 1):
 = integral2/((u^2 + 1) ((6 u)/(u^2 + 1) - (4 (1 - u^2))/(u^2 + 1))) du
Simplify the integrand 2/((u^2 + 1) ((6 u)/(u^2 + 1) - (4 (1 - u^2))/(u^2 + 1))) to get 1/(2 u^2 + 3 u - 2):
 = integral1/(2 u^2 + 3 u - 2) du
For the integrand 1/(2 u^2 + 3 u - 2), complete the square:
 = integral1/((sqrt(2) u + 3/(2 sqrt(2)))^2 - 25/8) du
For the integrand 1/((sqrt(2) u + 3/(2 sqrt(2)))^2 - 25/8), substitute s = sqrt(2) u + 3/(2 sqrt(2)) and ds = sqrt(2) du:
 = 1/sqrt(2) integral1/(s^2 - 25/8) ds
Factor -25/8 from the denominator:
 = 1/sqrt(2) integral-8/(25 (1 - (8 s^2)/25)) ds
Factor out constants:
 = -(4 sqrt(2))/25 integral1/(1 - (8 s^2)/25) ds
For the integrand 1/(1 - (8 s^2)/25), substitute p = (2 sqrt(2) s)/5 and dp = (2 sqrt(2))/5 ds:
 = -2/5 integral1/(1 - p^2) dp
The integral of 1/(1 - p^2) is tanh^(-1)(p):
 = -2/5 tanh^(-1)(p) + constant
Substitute back for p = (2 sqrt(2) s)/5:
 = -2/5 tanh^(-1)((2 sqrt(2) s)/5) + constant
Substitute back for s = sqrt(2) u + 3/(2 sqrt(2)):
 = -2/5 tanh^(-1)((4 u)/5 + 3/5) + constant
Substitute back for u = tan(x/2):
 = -2/5 tanh^(-1)(1/5 (4 tan(x/2) + 3)) + constant
Which is equivalent for restricted x values to:
Answer: | = 1/5 (log(cos(x/2) - 2 sin(x/2)) - log(sin(x/2) + 2 cos(x/2))) + constant

Guest Jun 16, 2017

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