Answer:
the answer is [tex]\frac{1}{41} (23x-2ln(5sinx+4cosx))[/tex]
Step-by-step explanation:
[tex]\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx[/tex]
here,
a=3, b=2, c=5, d=4
[tex]M=\frac{ac+bd}{c^2+d^2 }=\frac{3*5+2*4}{5^2+4^2}=\frac{23}{41}[/tex]
[tex]N=\frac{bc-ad}{c^2+d^2 }=\frac{2*5-3*4}{5^2+4^2}=\frac{-2}{41}[/tex]
[tex]\int {\frac{3sinx+2cosx}{5sinx+4cosx} } \, dx=\int{\frac{\frac{23}{41}(5sinx+4cosx)-\frac{2}{41}(5cosx-4sinx) }{5sinx+4cosx} }\,dx[/tex]
[tex]=\frac{23}{41}\int\, dx -\frac{2}{41} \int{\frac{5cosx-4sinx}{5sinx+4cosx} }\, dx[/tex]
[tex]=\frac{1}{41} (23x-2ln(5sinx+4cosx))[/tex]
pls mark it as the brainliest
