Problem to solve
Find the indefinite integral \[\int\frac{\sin[x]-3\cos[x]}{\cos[x]-2} dx\]
Solution
STEP 1
Take the integral: integral(sin(x) - 3 cos(x))/(cos(x) - 2) dx
STEP 2
For the integrand (sin(x) - 3 cos(x))/(cos(x) - 2), 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): = integral(2 ((2 u)/(u^2 + 1) - (3 (1 - u^2))/(u^2 + 1)))/((u^2 + 1) ((1 - u^2)/(u^2 + 1) - 2)) du
STEP 3
Simplify the integrand (2 ((2 u)/(u^2 + 1) - (3 (1 - u^2))/(u^2 + 1)))/((u^2 + 1) ((1 - u^2)/(u^2 + 1) - 2)) to get (-6 u^2 - 4 u + 6)/(3 u^4 + 4 u^2 + 1): = integral(-6 u^2 - 4 u + 6)/(3 u^4 + 4 u^2 + 1) du
STEP 4
For the integrand (-6 u^2 - 4 u + 6)/(3 u^4 + 4 u^2 + 1), use partial fractions: = integral(12 - 6 u)/(3 u^2 + 1) + (2 u - 6)/(u^2 + 1) du
STEP 5
Integrate the sum term by term: = integral(12 - 6 u)/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 6
For the integrand (12 - 6 u)/(3 u^2 + 1), cancel common terms in the numerator and denominator: = integral-(6 (u - 2))/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 7
Factor out constants: = -6 integral(u - 2)/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 8
Expanding the integrand (u - 2)/(3 u^2 + 1) gives u/(3 u^2 + 1) - 2/(3 u^2 + 1): = -6 integral(u/(3 u^2 + 1) - 2/(3 u^2 + 1)) du + integral(2 u - 6)/(u^2 + 1) du
STEP 9
Integrate the sum term by term and factor out constants: = -6 integral u/(3 u^2 + 1) du + 12 integral1/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 10
For the integrand u/(3 u^2 + 1), substitute s = 3 u^2 + 1 and ds = 6 u du: = - integral1/s ds + 12 integral1/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 11
The integral of 1/s is log(s): = -log(s) + 12 integral1/(3 u^2 + 1) du + integral(2 u - 6)/(u^2 + 1) du
STEP 12
For the integrand 1/(3 u^2 + 1), substitute p = sqrt(3) u and dp = sqrt(3) du: = -log(s) + 4 sqrt(3) integral1/(p^2 + 1) dp + integral(2 u - 6)/(u^2 + 1) du
STEP 13
The integral of 1/(p^2 + 1) is tan^(-1)(p): = 4 sqrt(3) tan^(-1)(p) - log(s) + integral(2 u - 6)/(u^2 + 1) du
STEP 14
For the integrand (2 u - 6)/(u^2 + 1), cancel common terms in the numerator and denominator: = 4 sqrt(3) tan^(-1)(p) - log(s) + integral(2 (u - 3))/(u^2 + 1) du
STEP 15
Factor out constants: = 4 sqrt(3) tan^(-1)(p) - log(s) + 2 integral(u - 3)/(u^2 + 1) du
STEP 16
Expanding the integrand (u - 3)/(u^2 + 1) gives u/(u^2 + 1) - 3/(u^2 + 1): = 4 sqrt(3) tan^(-1)(p) - log(s) + 2 integral(u/(u^2 + 1) - 3/(u^2 + 1)) du
STEP 17
Integrate the sum term by term and factor out constants: = 4 sqrt(3) tan^(-1)(p) - log(s) + 2 integral u/(u^2 + 1) du - 6 integral1/(u^2 + 1) du
STEP 18
For the integrand u/(u^2 + 1), substitute w = u^2 + 1 and dw = 2 u du: = 4 sqrt(3) tan^(-1)(p) - log(s) + integral1/w dw - 6 integral1/(u^2 + 1) du
STEP 19
The integral of 1/w is log(w): = 4 sqrt(3) tan^(-1)(p) - log(s) + log(w) - 6 integral1/(u^2 + 1) du
STEP 20
The integral of 1/(u^2 + 1) is tan^(-1)(u): = 4 sqrt(3) tan^(-1)(p) - log(s) - 6 tan^(-1)(u) + log(w) + constant
STEP 21
Substitute back for w = u^2 + 1: = 4 sqrt(3) tan^(-1)(p) - log(s) + log(u^2 + 1) - 6 tan^(-1)(u) + constant
STEP 22
Substitute back for p = sqrt(3) u: = -log(s) + log(u^2 + 1) - 6 tan^(-1)(u) + 4 sqrt(3) tan^(-1)(sqrt(3) u) + constant
STEP 23
Substitute back for s = 3 u^2 + 1: = log(u^2 + 1) - log(3 u^2 + 1) - 6 tan^(-1)(u) + 4 sqrt(3) tan^(-1)(sqrt(3) u) + constant
STEP 24
Substitute back for u = tan(x/2): = -3 x + 4 sqrt(3) tan^(-1)(sqrt(3) tan(x/2)) - log(3 tan^2(x/2) + 1) + log(sec^2(x/2)) + constant
STEP 25
An alternative form of the integral is: = -3 x + 4 sqrt(3) tan^(-1)(sqrt(3) tan(x/2)) + log(1/(2 - cos(x))) + constant
STEP 26
Which is equivalent for restricted x values to: Answer: | | = -3 x + 4 sqrt(3) tan^(-1)(sqrt(3) tan(x/2)) - log(2 - cos(x)) + constant
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Edited by tododemate® - 2021