Complete the square, then do a trigonometric substitution.
sqrt(13+12x-x^2)=sqrt(49-36+12x-x^2)=sqrt(49-(x-6)^2)√13+12x−x2=√49−36+12x−x2=√49−(x−6)2
So, you want to evaluate intsqrt(49-(x-6)^2)dx∫√49−(x−6)2dx.
It's going to get very messy, so I want to let x-6=7ux−6=7u. It is then easy to see that (x-6)^2=49u^2(x−6)2=49u2 and dx=7dudx=7du, so the integral becomes int7sqrt(49-49u^2) du=7int7sqrt(1-u^2)du=49intsqrt(1-u^2)du∫7√49−49u2du=7∫7√1−u2du=49∫√1−u2du,
Now do a trigonometric substitution. I'll use u=sinthetau=sinθ which, of course, makes du=costheta d thetadu=cosθdθ.
This makes the integral: 49intsqrt(1-sin^2theta)costhetad theta49∫√1−sin2θcosθdθ.
This, in turn, becomes 49intcos^2thetad theta49∫cos2θdθ. Now, use the power reduction to re-write cos^2thetacos2θ as 1/2(1+cos2theta)12(1+cos2θ).
The problem has become:
Evaluate: 49/2int(1+cos2theta)d theta=49/2(theta+1/2sin2theta)+C492∫(1+cos2θ)dθ=492(θ+12sin2θ)+C
Backfilling: theta=sin^-1uθ=sin−1u and
since u=sinthetau=sinθ implies costheta=sqrt(1-u^2)cosθ=√1−u2
1/2sin2theta=1/2(2sinthetacostheta)=usqrt(1-u^2)12sin2θ=12(2sinθcosθ)=u√1−u2
Thus, the integral evaluates(in terms of uu) leaving +C for later, to: 49/2(sin^-1u+usqrt(1-u^2))=1/2(49sin^-1u+49usqrt(1-u^2))=1/2(49sin^-1((x-6)/7)+49((x-6)/7)sqrt(1-((x-6)/7)^2))=1/2(49sin^-1((x-6)/7)+49((x-6)/7)sqrt(1-((x-6)/7)^2))=1/2(49sin^-1((x-6)/7)+(x-6)sqrt(49-(x-6)^2))=1/2(49sin^-1((x-6)/7)+(x-6)sqrt(13+12x-x^2))
Oh, yeah! Don't forget to put +C.
