The answer is:
#49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c#.
Follow my passages:
#intsqrt(13+12x-x^2)dx=intsqrt(13-(x^2-12x))dx=#
#=intsqrt(13-(x^2-12x+36-36))dx=intsqrt(13+36-(x-6)^2)dx=#
#=intsqrt(49-(x-6)^2)dx=(1)#.
Since
#x-6=7sintrArrx=7sint+6rArrdx=7costdt#,
our integral becomes:
#(1)=intsqrt(49-(7sint)^2)*7costdt=#
#=intsqrt(49-49sin^2t)*7costdt=#
#=intsqrt(49(1-sin^2t))*7costdt=int7sqrt(1-sin^2t)*7costdt=#
#=49intsqrt(cos^2t)*costdt=49intcos^2tdt=(2)#.
Since:
#cos(alpha/2)=+-sqrt((1+cosalpha)/2#,
our integral becomes:
#(2)=49int(1+cos2t)/2dt=49/2(intdt+1/2int2cos2tdt)=#
#=49/2(t+1/2sin2t)+c=49/2t+49/2*1/2*2sintcost+c=#
#=49/2t+49/2sintcost+c=(3)#.
Since
#sint=(x-6)/7rArrt=arcsin((x-6)/7)#
and
#cost=sqrt(1-sin^2t)=sqrt(1-((x-6)/7)^2)=#
#=sqrt(1-(x^2-12x+36)/49)=sqrt((49-x^2+12x-36)/49=#
#=sqrt(13+12x-x^2)/7#,
our integral becomes:
#(3)=49/2arcsin((x-6)/7)+49/2*(x-6)/7*sqrt(13+12x-x^2)/7+c=#
#=49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c#.
