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Answer 1 · 2017-12-13T19:11:24

#int ((2sinx+cosx)*dx)/(7sinx-5cosx)#=#9/74*x+17/74Ln(7sinx-5cosx)+C#

I used #2sinx+cosx=a*(7sinx-5cosx)+b*(7cosx+5sinx)# equation for calculating this integral,

#(7a+5b)*sinx+(7b-5a)*cosx=2sinx+cosx#

After equating coefficients, #7a+5b=2# and #7b-5a=1#

After solving these equations, I found #a=9/74# and #b=17/74#

Thus,

#int ((2sinx+cosx)*dx)/(7sinx-5cosx)#

=#9/74*int ((7sinx-5cosx)*dx)/(7sinx-5cosx)#+#17/74*int ((7cosx+5sinx)*dx)/(7sinx-5cosx)#

=#9/74*int dx#+#17/74*int ((7cosx+5sinx)*dx)/(7sinx-5cosx)#

=#9/74*x+17/74Ln(7sinx-5cosx)+C#