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[SOLVED] Question about Trigonometric Substitution

MacLaddy

Member
Jan 29, 2012
52
Hello again,

I have a question about a trigonometric substitution problem that I am struggling with. I was able to get the correct answer, which I know is correct because of Wolfram verification, and my school has a way of showing an example which showed the steps.... Anyway, see below.

\(\displaystyle \int_0^\frac{1}{5}\frac{dx}{\sqrt{25x^2+1}^\frac{3}{2}}\)

Without going through all the steps, I can tell you that the solution to this integral is,

\(\displaystyle [\frac{1}{5}sin\theta]_0^\frac{1}{5}\)

which SHOULD simplify to

\(\displaystyle [\frac{1}{5}*\frac{x}{\sqrt{25x^2+1}}]_0^\frac{1}{5}\)

However, the book shows that \(\displaystyle \frac{1}{5}\) is omitted, and this ends up being the final correct answer by taking 1/5 out.

Can anyone explain to me what is happening here?

Thanks,
Mac
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
I believe when you set up the triangle for trig substitution you have the legs as 5x and 1, thus the hypotenuse is \(\displaystyle \sqrt{25x^2+1}\). If I'm not switching the orientation of the legs then that means \(\displaystyle \sin(\theta)=\frac{5x}{\sqrt{25x^2+1}}\) so you forgot a 5.
 

MacLaddy

Member
Jan 29, 2012
52
Ah, of course. I tried that problem repeatedly and continued making the same mistake.

Thanks again, Jameson.