November 4th, 2014, 03:46 PM  #1 
Newbie Joined: Oct 2014 From: California Posts: 5 Thanks: 0  Integral with trig identity (help)
$\displaystyle \int \sqrt{40+6xx^2}dx$ So I solved it out and managed to get the answer using (x3) = 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!!! 
November 4th, 2014, 04:17 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,887 Thanks: 1506 
I graphed both on my TI83 (assuming C = 0 for both) ... their graphs coincide.


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identity, integral, trig 
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