Solving the Indefinite Integral: \int\sqrt{28x-x^2} dx

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Homework Statement



Evaluate the indefinite integral.

\int\sqrt{28x-x^2} dx

The Attempt at a Solution



\int\sqrt{196-(x-14)^2} dx
Completing the square​

u=x-14

du=dx

\int\sqrt{196-u^2} du
u substitution​

u=14sin\theta

du=14cos\theta d\theta
Trig substitution​

\int\sqrt{196cos^2\theta} 14cos\theta d\theta

\int14cos\theta*14cos\theta d\theta

98\int1+cos2\theta d\theta

98(\theta+sin\theta*cos\theta) + C

98(arcsin(u/14)+(u/14)(\sqrt{196-u^2}/14)
Solve for Theta​

From here I should be able to just sub in for u and arrive at my answer. Unfortunately, it is close but no correct, and I can't see where I'm going wrong. Any ideas?
 
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What's the answer you're getting and what's the answer you're supposed to get. I don't see any errors in what you've written, but perhaps its a simplification error.
 
Thanks for the prompt replies :)

So it appears my problem is either data entry or substitution; that's a good sign. Here is my final solution exactly as I've entered it in our software:

98(arcsin((x-14)/14)+(x-14)sqrt(x^2-28x)/2)

Sorry if that's not to pretty; here's the image of the same input:

[PLAIN]http://webwork.asu.edu/webwork2_files/tmp/equations/a6/bd057be8bc635789c21be6e19460f51.png

Comparing it to the Wolfram output, it looks close but not identical. Then again, the computer tends to do some crazy simplification at the end; maybe they're equivalent. See: http://www.wolframalpha.com/input/?i=integrate+sqrt(28x-x^2)
 
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talk2glenn said:
Thanks for the prompt replies :)

So it appears my problem is either data entry or substitution; that's a good sign. Here is my final solution exactly as I've entered it in our software:

98(arcsin((x-14)/14)+(x-14)sqrt(x^2-28x)/2)

Doesn't that last term need to be

\frac{(x-14)\sqrt{28x-x^2}}{196}

So you need to switch the entries in your root around. and I don't really see why you only divide by 2.

Sorry if that's not to pretty; here's the image of the same input:

[PLAIN]http://webwork.asu.edu/webwork2_files/tmp/equations/a6/bd057be8bc635789c21be6e19460f51.png

Comparing it to the Wolfram output, it looks close but not identical. Then again, the computer tends to do some crazy simplification at the end; maybe they're equivalent. See: http://www.wolframalpha.com/input/?i=integrate+sqrt(28x-x^2)

The wolfram output is correct. However, wolfram uses an entirely different algorithm to calculate this. So the answers can look pretty different...
 
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You're right; I see my error now. On paper, I distributed the 98*stuff/196 and got stuff/2, but when I entered it I kept the 98 as a factor. Edit: And you're right about the order too- I reversed it. Always rushing these things at the end.

Let me try that and see what happens. Thanks again!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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