`int(3 sinx+ 2 cos x)/(3 cosx+2 sin x)dx= ax+b log |3cos x + 2 sin x| +C`, then (a ,b)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)`

`int(3 sinx+ 2 cos x)/(3 cosx+2 sin x)dx= ax+b log |3cos x + 2 sin x|

1.	`int(3 sinx+ 2 cos x)/(3 cosx+2 sin x)dx= ax+b log \|3cos x + 2 sin x\| +C`, then (a ,b)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» Correct Answer - b 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)`

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