1
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$= \int \dfrac{1}{\sqrt{(5y-1)^2-4}}dy$

$=\int \dfrac{1}{\sqrt{u^2-4}}\dfrac{du}{5}, \quad U$ substitution

$=\int \dfrac{1}{10\cos(\theta)} 2\cos(\theta) d\theta, \quad$ Trig substitution

$= \dfrac{1}{5} \theta$

$= \dfrac{\cos^{-1}\frac{\sqrt{(5y-1)^2-4}}{2}}{5} +C$

Where did I go wrong?

work after u substitution $\frac{1}{5}\int \sec \theta d\theta \\ =\frac{1}{5} \ln \left | \sec \theta + \tan \theta\right | $

* New Answer* $\\ \frac{1}{5}\ln \left | \frac{5y-1}{2} +\frac{\sqrt{(5y-1)^2 -4}}{2}\right | +C$

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  • $\begingroup$ Uh-you sure it's wrong? lol Seriously,though-how'd you get the last step-a geometric definition of theta? Sorry,I'm tired and it's not clear to me where you got it. $\endgroup$ Jan 31 '15 at 2:36
  • $\begingroup$ @Mathemagician1234 I just edited it the 2 wasn't supposed to be in the square root, but i had $u=2sin\theta$ so I made a triangle off that. $\endgroup$
    – Sean
    Jan 31 '15 at 2:40
  • $\begingroup$ @Tyler Since you have an irreducible quadratic, a better method might be to complete the square and then perform a trig substitution. $\endgroup$
    – user156926
    Jan 31 '15 at 2:41
  • $\begingroup$ Looks ok to me,but it IS a pretty gnarly computation,so you really would like to check it with a computer algebra program. $\endgroup$ Jan 31 '15 at 2:41
  • $\begingroup$ @Jun-GooKwak Thats what I did $\endgroup$
    – Sean
    Jan 31 '15 at 2:42
0
$\begingroup$

The problem is in the 3rd line. $$\sqrt{4\sin^2\theta-4} \neq 2\cos \theta$$ Try $x=2\cosh\theta$ or $x=2\sec\theta$

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14
  • $\begingroup$ ahh would it be $-cos\theta$ $\endgroup$
    – Sean
    Jan 31 '15 at 2:46
  • $\begingroup$ No it would not. $1-\sin^2\theta= -\cos^2\theta$, but you cannot take a complex root. $\endgroup$
    – Teoc
    Jan 31 '15 at 2:46
  • $\begingroup$ @MathNoob Ah,were it only that simple. $\endgroup$ Jan 31 '15 at 2:47
  • $\begingroup$ i mean using the formula $\sin^2\theta + \cos^2\theta = 1$ $\endgroup$
    – Sean
    Jan 31 '15 at 2:47
  • $\begingroup$ $\sin^2\theta - 1 = -\cos^2\theta$ $\endgroup$
    – Sean
    Jan 31 '15 at 2:48

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