How do you find the integral of sqrt(13+12x-x^2)dx13+12xx2dx?

1 Answer
Feb 21, 2015

The answer is:

49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c492arcsin(x67)+x6213+12xx2+c.

Follow my passages:

intsqrt(13+12x-x^2)dx=intsqrt(13-(x^2-12x))dx=13+12xx2dx=13(x212x)dx=

=intsqrt(13-(x^2-12x+36-36))dx=intsqrt(13+36-(x-6)^2)dx==13(x212x+3636)dx=13+36(x6)2dx=

=intsqrt(49-(x-6)^2)dx=(1)=49(x6)2dx=(1).

Since

x-6=7sintrArrx=7sint+6rArrdx=7costdtx6=7sintx=7sint+6dx=7costdt,

our integral becomes:

(1)=intsqrt(49-(7sint)^2)*7costdt=(1)=49(7sint)27costdt=

=intsqrt(49-49sin^2t)*7costdt==4949sin2t7costdt=

=intsqrt(49(1-sin^2t))*7costdt=int7sqrt(1-sin^2t)*7costdt==49(1sin2t)7costdt=71sin2t7costdt=

=49intsqrt(cos^2t)*costdt=49intcos^2tdt=(2)=49cos2tcostdt=49cos2tdt=(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)=491+cos2t2dt=492(dt+122cos2tdt)=

=49/2(t+1/2sin2t)+c=49/2t+49/2*1/2*2sintcost+c==492(t+12sin2t)+c=492t+492122sintcost+c=

=49/2t+49/2sintcost+c=(3)=492t+492sintcost+c=(3).

Since

sint=(x-6)/7rArrt=arcsin((x-6)/7)sint=x67t=arcsin(x67)

and

cost=sqrt(1-sin^2t)=sqrt(1-((x-6)/7)^2)=cost=1sin2t=1(x67)2=

=sqrt(1-(x^2-12x+36)/49)=sqrt((49-x^2+12x-36)/49==1x212x+3649=49x2+12x3649=

=sqrt(13+12x-x^2)/7=13+12xx27,

our integral becomes:

(3)=49/2arcsin((x-6)/7)+49/2*(x-6)/7*sqrt(13+12x-x^2)/7+c=(3)=492arcsin(x67)+492x6713+12xx27+c=

=49/2arcsin((x-6)/7)+(x-6)/2sqrt(13+12x-x^2)+c=492arcsin(x67)+x6213+12xx2+c.