Let, #I=int3/(4+5sinx)dx#.
The substn. to be used for this type is #tan(x/2)=t#.
#:. (sec^2(x/2))(1/2)dx=dt, or, dx=(2dt)/(sec^2(x/2)), i.e., #
#dx=(2dt)/(1+tan^2(x/2))=(2dt)/(1+t^2)#.
Further, #4+5sinx=4+5{(2tan(x/2))/(1+tan^2(x/2))}#,
#=4+5{(2t)/(1+t^2)}#,
#:. 4+5sinx=(4t^2+10t+4)/(1+t^2)={2(2t^2+5t+2)}/(1+t^2)#.
#:. I=3int1/(4+5sinx)dx#,
#=3int(1+t^2)/{2(2t^2+5t+2)}*2/(1+t^2)dt#,
#=3int1/(2t^2+5t+2)dt#,
#=int3/{(t+2)(2t+1)dt#,
#=int3/{(t+2)2(t+1/2)}dt#,
#=int(3/2)/{(t+2)(t+1/2)}dt#,
#=int{(t+2)-(t+1/2)}/{(t+2)(t+1/2)}dt#,
#=int{(t+2)/{(t+2)(t+1/2)}-(t+1/2)/{(t+2)(t+1/2)}}dt#,
#=int{1/(t+1/2)-1/(t+2)}dt#,
#=ln|(t+1/2)|-ln|(t+2)|#.
Replacing #t" by "tan(x/2)#, we have,
# I=ln|(2tan(x/2)+1)/(2(tan(x/2)+2))|+C, #
# or, I=ln|(2sin(x/2)+cos(x/2))/(2(sin(x/2)+2cos(x/2)))|+C#.
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