The answer is:
49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c492arcsin(x−67)+x−62√13+12x−x2+c.
Follow my passages:
intsqrt(13+12x-x^2)dx=intsqrt(13-(x^2-12x))dx=∫√13+12x−x2dx=∫√13−(x2−12x)dx=
=intsqrt(13-(x^2-12x+36-36))dx=intsqrt(13+36-(x-6)^2)dx==∫√13−(x2−12x+36−36)dx=∫√13+36−(x−6)2dx=
=intsqrt(49-(x-6)^2)dx=(1)=∫√49−(x−6)2dx=(1).
Since
x-6=7sintrArrx=7sint+6rArrdx=7costdtx−6=7sint⇒x=7sint+6⇒dx=7costdt,
our integral becomes:
(1)=intsqrt(49-(7sint)^2)*7costdt=(1)=∫√49−(7sint)2⋅7costdt=
=intsqrt(49-49sin^2t)*7costdt==∫√49−49sin2t⋅7costdt=
=intsqrt(49(1-sin^2t))*7costdt=int7sqrt(1-sin^2t)*7costdt==∫√49(1−sin2t)⋅7costdt=∫7√1−sin2t⋅7costdt=
=49intsqrt(cos^2t)*costdt=49intcos^2tdt=(2)=49∫√cos2t⋅costdt=49∫cos2tdt=(2).
Since:
cos(alpha/2)=+-sqrt((1+cosalpha)/2cos(α2)=±√1+cosα2,
our integral becomes:
(2)=49int(1+cos2t)/2dt=49/2(intdt+1/2int2cos2tdt)=(2)=49∫1+cos2t2dt=492(∫dt+12∫2cos2tdt)=
=49/2(t+1/2sin2t)+c=49/2t+49/2*1/2*2sintcost+c==492(t+12sin2t)+c=492t+492⋅12⋅2sintcost+c=
=49/2t+49/2sintcost+c=(3)=492t+492sintcost+c=(3).
Since
sint=(x-6)/7rArrt=arcsin((x-6)/7)sint=x−67⇒t=arcsin(x−67)
and
cost=sqrt(1-sin^2t)=sqrt(1-((x-6)/7)^2)=cost=√1−sin2t=√1−(x−67)2=
=sqrt(1-(x^2-12x+36)/49)=sqrt((49-x^2+12x-36)/49==√1−x2−12x+3649=√49−x2+12x−3649=
=sqrt(13+12x-x^2)/7=√13+12x−x27,
our integral becomes:
(3)=49/2arcsin((x-6)/7)+49/2*(x-6)/7*sqrt(13+12x-x^2)/7+c=(3)=492arcsin(x−67)+492⋅x−67⋅√13+12x−x27+c=
=49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c=492arcsin(x−67)+x−62√13+12x−x2+c.