1.

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

Answer» Correct Answer - C::D
Differentiating both sides w.r.t. x, we get
`(3sin x+2 cosx)/(3cosx+2sinx)=a+(b(2cosx-3sinx))/((2sinx+3cosx))`
`=(sinx.(2a-3b)+cosx.(3a+2b))/((3cosx+2 sinx))`
Comparing like terms on both sides, we get
`3=2a-3b, 2=3a+2b`
`implies a=(12)/(13), b= -(15)/(39)`


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