If ∫3sin⁡x+2cos⁡x3cos⁡x+2sin⁡xdx=ax+bln⁡2sinx+3cosx+C then

Differentiating both sides, we get3sin⁡x+2cos⁡x3cos⁡x+2sin⁡x=a+b(2cos⁡x−3sin⁡x)(2sin⁡x+3cos⁡x)=sin⁡x⋅(2a−3b)+cos⁡x⋅(3a+2b)(3cos⁡x+2sin⁡x)Comparing like terms on both sides, we get 3=2a−3b,2=3a+2b⇒a=1213,b=−1539