# integrate:(6+ 3sin x + 14cos x) / (3+4sin x+5cos x)

Vikas TU
14149 Points
7 years ago
Dear Student,
integral (3 sin(x) + 14 cos(x) + 6)/(4 sin(x) + 5 cos(x) + 3) dx
For the integrand (3 sin(x) + 14 cos(x) + 6)/(4 sin(x) + 5 cos(x) + 3),
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 ((6 u)/(u^2 + 1) + (14 (1 - u^2))/(u^2 + 1) + 6))/((u^2 + 1) ((8 u)/(u^2 + 1) + (5 (1 - u^2))/(u^2 + 1) + 3)) du
Simplify the integrand (2 ((6 u)/(u^2 + 1) + (14 (1 - u^2))/(u^2 + 1) + 6))/((u^2 + 1) ((8 u)/(u^2 + 1) + (5 (1 - u^2))/(u^2 + 1) + 3)) to get (8 u^2 - 6 u - 20)/(u^4 - 4 u^3 - 3 u^2 - 4 u - 4):
=integral (8 u^2 - 6 u - 20)/(u^4 - 4 u^3 - 3 u^2 - 4 u - 4) du
For the integrand (8 u^2 - 6 u - 20)/(u^4 - 4 u^3 - 3 u^2 - 4 u - 4), partial fractions:
=integral ((4 - 2 u)/(u^2 + 1) + 1/(u - 2 sqrt(2) - 2) + 1/(u + 2 sqrt(2) - 2)) du
Integrating term by term:
=integral (4 - 2 u)/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
=integral -(2 (u - 2))/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
=-2 integral (u - 2)/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
Expanding the integrand (u - 2)/(u^2 + 1) gives u/(u^2 + 1) - 2/(u^2 + 1):
=-2 integral (u/(u^2 + 1) - 2/(u^2 + 1)) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
Integrate term by term and factor out constant:
=-2 integral u/(u^2 + 1) du + 4 integral 1/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
For u/(u^2 + 1), substitute s = u^2 + 1 and  ds = 2 u  du:
=- integral 1/s ds + 4 integral 1/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
The integral of 1/s is log(s):
=-log(s) + 4 integral 1/(u^2 + 1) du +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
The integral of 1/(u^2 + 1) is tan^(-1)(u):
=4 tan^(-1)(u) - log(s) +  integral 1/(u + 2 sqrt(2) - 2) du +  integral 1/(u - 2 sqrt(2) - 2) du
For 1/(u + 2 sqrt(2) - 2), substitute p = u + 2 sqrt(2) - 2 and  dp =du:
=4 tan^(-1)(u) - log(s) +  integral 1/p dp +  integral 1/(u - 2 sqrt(2) - 2) du
The integral of 1/p is log(p):
=4 tan^(-1)(u) + log(p) - log(s) +  integral 1/(u - 2 sqrt(2) - 2) du
For the integrand 1/(u - 2 sqrt(2) - 2), substitute w = u - 2 sqrt(2) - 2 and  dw =du:
=4 tan^(-1)(u) + log(p) - log(s) +  integral 1/w dw
The integral of 1/w is log(w):
=log(p) - log(s) + 4 tan^(-1)(u) + log(w) + constant
Substitute back for w = u - 2 sqrt(2) - 2:
=log(p) - log(s) + log(u - 2 sqrt(2) - 2) + 4 tan^(-1)(u) + constant
Substitute back for p = u + 2 sqrt(2) - 2:
=-log(s) + log(u - 2 sqrt(2) - 2) + log(u + 2 sqrt(2) - 2) + 4 tan^(-1)(u) + constant
Substitute back for s = u^2 + 1:
=-log(u^2 + 1) + log(u - 2 sqrt(2) - 2) + log(u + 2 sqrt(2) - 2) + 4 tan^(-1)(u) + constant
Substitute back for u = tan(x/2):
=2 x + log(tan(x/2) - 2 sqrt(2) - 2) + log(tan(x/2) + 2sqrt(2)- 2) - log(sec^2(x/2)) + constant
can be written as,
=2 x + log(1/2 (-4 sin(x) - 5 cos(x) - 3)) + c
so ans=2 x + log(4 sin(x) + 5 cos(x) + 3) + c.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)