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Q.
$\int\frac{ 3 \,sin \,x + 2 \,cos\, x}{3 \,cos\, x + 2 \,sin\, x} dx = ax + b\, ln\, \left| 2 \,sin \,x + 3\, cos \,x \right| + C $ then
Integrals
Solution:
Differentiating both sides, we get
$\frac{3 \,sin \,x + 2 \,cos \,x}{3 \,cos \,x + 2 \,sin \,x} = a + \frac{b\left(2 \, cos \, x - 3 \,sin \, x\right)}{\left(2 \,sin \,x + 3 \, cos \, x\right)} $
$ = \frac{sin \, x\left(2a - 3b\right)+ cos \, x \left( 3a + 2b\right)}{\left(3 \,cos \, x + 2 \, sin x\right)}$
Comparing like terms on both sides, we get$3 = 2a-3b, 2 - 3a + 2b$
or $a=\frac{12}{13}, b=-\frac{15}{39}$