How do I evaluate intsqrt(13+12x-x^2)dx13+12xx2dx?

1 Answer
Mar 6, 2015

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+12xx2=4936+12xx2=49(x6)2

So, you want to evaluate intsqrt(49-(x-6)^2)dx49(x6)2dx.

It's going to get very messy, so I want to let x-6=7ux6=7u. It is then easy to see that (x-6)^2=49u^2(x6)2=49u2 and dx=7dudx=7du, so the integral becomes int7sqrt(49-49u^2) du=7int7sqrt(1-u^2)du=49intsqrt(1-u^2)du74949u2du=771u2du=491u2du,

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 theta491sin2θcosθdθ.

This, in turn, becomes 49intcos^2thetad theta49cos2θdθ. 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=sin1u and
since u=sinthetau=sinθ implies costheta=sqrt(1-u^2)cosθ=1u2

1/2sin2theta=1/2(2sinthetacostheta)=usqrt(1-u^2)12sin2θ=12(2sinθcosθ)=u1u2

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.