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November 4th, 2014, 03:46 PM   #1
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Integral with trig identity (help)

$\displaystyle \int \sqrt{40+6x-x^2}dx$

So I solved it out and managed to get the answer using (x-3) = 7sin(theta)

$\displaystyle \frac{49}{2}(\theta+\frac{sin(2\theta)}{2})+C$

However, using the double angle identity, does the answer above equal:

$\displaystyle \frac{49}{2}(\theta+sin\theta cos\theta)+C$

I punched both equations to my calculator and Symbolab a bunch of times and the definite answers seem to be different. I must be missing something .

Thanks in advance!!!
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November 4th, 2014, 04:17 PM   #2
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I graphed both on my TI-83 (assuming C = 0 for both) ... their graphs coincide.
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