Question

Evaluate it..........​

Answer 1

Answer:

the answer is $\frac{1}{41} (23x-2ln(5sinx+4cosx))$

Step-by-step explanation:

$\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx$

here,

a=3, b=2, c=5, d=4

$M=\frac{ac+bd}{c^2+d^2 }=\frac{3*5+2*4}{5^2+4^2}=\frac{23}{41}$

$N=\frac{bc-ad}{c^2+d^2 }=\frac{2*5-3*4}{5^2+4^2}=\frac{-2}{41}$

$\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx=\int{\frac{\frac{23}{41}(5sinx+4cosx)-\frac{2}{41}(5cosx-4sinx) }{5sinx+4cosx} }\,dx$

$=\frac{23}{41}\int\, dx -\frac{2}{41} \int{\frac{5cosx-4sinx}{5sinx+4cosx} }\, dx$

$=\frac{1}{41} (23x-2ln(5sinx+4cosx))$

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