If `int (3sin x+2 cosx)/(3cosx+2sinx)dx=ax+blog_(e)|2 sinx+3 cosx|+c` then a. `a= -(12)/(13)` b. `b=(6)/(13)` c. `a=(12)/(13)` d. `b= -(15)/(39)`

Options

A. `a=(5)/(13), b =-(12)/(13)`
B. `a=(12)/(13),b =-(5)/(13)`
C. `a=(12)/(13),b=(5)/(13)`
D. `a=(-12)/(5),b=(-5)/(13)`

Answer

b

Explanation

Let `I= int(3 sin x+2 cos x)/(3 cos x+2sinx)dx`
Let `3 sin x +2 cos x= lambda (d)/(dx)(3cosx +2sinx)+mu(3 cos x +2 sinx)`
`rArr 3 sin x +2 cos x = lambda(-3 sin x +2 cos x )+mu (3 cos x+2 sin x)`
Comparing the coefficients sinx and cos x on both sides , we get
`-3lambda_2mu = 3 and 2 lambda +3 mu =2`
` rArr lambda =-5 //13 and = 12//13`
`:. I= int(-((5)/(13))(-3sin x +2 cos x)+((12)/(13))(3 cos x +2 sin x))/(3 cos x+ 2 sin x)dx`
`rArr I=(12)/(13)int1*dx - (5)/(13) int (1)/(3 cos x +2sin x)d(3cos x +2 sin x)`
`rArr I=(12)/(13) x-(5)/(13)log |3 cos x+2 sin x| +C`
Hence , a `=(12)/(13) and b = (-5)/(13)`