MHB Integral of [1/sqrt(8x-x^2)] dx - Daisy Dee's Answer at Yahoo Answers

[MarkFL]

Here is the question:

Integrate [1/sqrt of (8x-x^2)] dx?

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Re: Daisy Dee's question: evaluating a definite integral where a trig. substitution is appropriate

Hello Daisy Dee,

We are given to evaluate:

$$I=\int\frac{1}{\sqrt{8x-x^2}}\,dx$$

We find that the domain of the integrand is $(0,8)$.

Completing the square on the radicand, we obtain:

$$I=\int\frac{1}{\sqrt{4^2-(x-4)^2}}\,dx$$

Using the substitution:

$$x-4=4\sin(\theta)\,\therefore\,dx=4 \cos(\theta)\,d\theta$$

We may write:

$$I=\int\frac{4\cos(\theta)}{\sqrt{4^2-4^2\sin^2(\theta)}}\,d\theta$$

For $$-\frac{\pi}{2}<\theta<\frac{\pi}{2}$$ we may write:

$$I=\int\,d\theta=\theta+C$$

Back substituting for $\theta$, we obtain:

$$I=\sin^{-1}\left(\frac{x-4}{4} \right)+C$$