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i’m not sure ‘bout the answer although i slved it so please do helh me ….integration of sin2x/(1+sinx)(2+six)*dx

i’m not sure ‘bout the answer although i slved it so please do helh me ….integration of sin2x/(1+sinx)(2+six)*dx

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
Given integral:
 integral (sin(2 x))/((sin(x) + 1) (sin(x) + 2)) dx
Using double angle formula sin(2 x) = 2 sin(x) cos(x):
  =integral (2 sin(x) cos(x))/((sin(x) + 1) (sin(x) + 2)) dx
Rewrite (2 sin(x) cos(x))/((sin(x) + 1) (sin(x) + 2)) as (2 sin(x) cos(x))/(2 + 3 sin(x) + sin^2(x)):
  =integral (2 sin(x) cos(x))/(2 + 3 sin(x) + sin^2(x)) dx
For the integrand (2 sin(x) cos(x))/(2 + 3 sin(x) + sin^2(x)), substitute u = sin(x) and  du = cos(x)  dx:
  =integral (2 u)/(u^2 + 3 u + 2) du
Factor out constants:
  =2 integral u/(u^2 + 3 u + 2) du
Rewrite the integrand u/(u^2 + 3 u + 2) as (2 u + 3)/(2 (u^2 + 3u + 2)) - 3/(2 (u^2 + 3 u + 2)):
  =2 integral ((2 u + 3)/(2 (u^2 + 3 u + 2)) - 3/(2 (u^2 + 3 u + 2))) du
Integrate the sum term by term and factor out constants:
  =integral (2 u + 3)/(u^2 + 3 u + 2) du - 3 integral 1/(u^2 + 3 u + 2) du
For the integrand (2 u + 3)/(u^2 + 3 u + 2), substitute s = u^2 + 3 u + 2 and  ds = (2 u + 3)  du:
  =integral 1/s ds - 3 integral 1/(u^2 + 3 u + 2) du
The integral of 1/s is log(s):
  log(s) - 3 integral 1/(u^2 + 3 u + 2) du
For the integrand 1/(u^2 + 3 u + 2), complete the square:
  =log(s) - 3 integral 1/((u + 3/2)^2 - 1/4) du
For the integrand 1/((u + 3/2)^2 - 1/4), substitute p = u + 3/2 and  dp =du:
  =log(s) - 3 integral 1/(p^2 - 1/4) dp
Factor -1/4 from the denominator:
  =log(s) - 3 integral 4/(4 p^2 - 1) dp
Factor out constants:
  =log(s) - 12 integral 1/(4 p^2 - 1) dp
Factor -1 from the denominator:
  =log(s) + 12 integral 1/(1 - 4 p^2) dp
For the integrand 1/(1 - 4 p^2), substitute w = 2 p and  dw = 2  dp:
  =log(s) + 6 integral 1/(1 - w^2) dw
The integral of 1/(1 - w^2) is tanh^(-1)(w):
  =log(s) + 6 tanh^(-1)(w) + constant
Substitute back for w = 2 p:
  =6 tanh^(-1)(2 p) + log(s) + constant
Substitute back for p = u + 3/2:
  =log(s) + 6 tanh^(-1)(2 u + 3) + constant
Substitute back for s = u^2 + 3 u + 2:
  =log(u^2 + 3 u + 2) + 6 tanh^(-1)(2 u + 3) + c
Substitute back for u = sin(x):
  =log(sin^2(x) + 3 sin(x) + 2) + 6 tanh^(-1)(2 sin(x) + 3) + c
Which is equal to:
Ans=4 log(sin(x) + 2) - 4 log(sin(x/2) + cos(x/2)) + c
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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