The integral is
#I=int((2cosx+7)dx)/(4-sinx)=int(2cosxdx)/(4-sinx)+int(7dx)/(4-sinx)#
#=I_1+I_2#
To calculate #I_1#, perform the substitution
#u=4-sinx#, #=>#, #du=-cosxdx#
Therefore,
#I_1=2int(cosxdx)/(4-sinx)=-2int(du)/(u)#
#=-2ln(u)#
#=-2ln(4-sinx)#
To calculate #I_2#, apply the Weierstrass substitution,
#sinx=(2tan(x/2))/(1+tan^2(x/2))#
Therefore,
#I_2=int(7dx)/(4-sinx)=7int(dx)/(4-((2tan(x/2))/(1+tan^2(x/2))))#
#=7int(1+tan^2(x/2)dx)/(4+4tan^2(x/2)-2tan(x/2))#
Let #t=tan(x/2)#, #=>#, #dt=1/2sec^2(x/2)dx=1/2(1+tan^2(x/2))dx=1/2(1+t^2)dx#
Therefore,
#I_2=14int(dt)/(4t^2-2t+4)=7int(dt)/(2t^2-t+2)#
Complete the square in the denominator
#2t^2-t+2=2(t^2-t/2+1)=2(t^2-t/2+1/16+1-1/16)#
#=2((t-1/4)^2+15/16)#
#I_2=7/2int(dt)/((t-1/4)^2+15/16)#
#=7/2int(dt)/((sqrt2t-1/2^(3/2))^2+15/8)#
Let #u=(4t-1)/sqrt15#, #=>#, #du=4/sqrt15dt#
#I_2=14int(sqrt15du)/(15u^2+15)#
#=14/sqrt15int(du)/(u^2+1)#
#=14/sqrt15arctan(u)#
#=14/sqrt15arctan((4t-1)/sqrt15)#
#=14/sqrt15arctan((4tan(x/2)-1)/sqrt15)#
Finally,
#I=14/sqrt15arctan((4tan(x/2)-1)/sqrt15)-2ln(4-sinx)+C#
