MHB Question about Trigonometric Substitution

[MacLaddy1]

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.

[math]\int_0^\frac{1}{5}\frac{dx}{\sqrt{25x^2+1}^\frac{3}{2}}[/math]

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

[math][\frac{1}{5}sin\theta]_0^\frac{1}{5}[/math]

which SHOULD simplify to

[math][\frac{1}{5}*\frac{x}{\sqrt{25x^2+1}}]_0^\frac{1}{5}[/math]

However, the book shows that [math]\frac{1}{5}[/math] 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]

I believe when you set up the triangle for trig substitution you have the legs as 5x and 1, thus the hypotenuse is [math]\sqrt{25x^2+1}[/math]. If I'm not switching the orientation of the legs then that means [math]\sin(\theta)=\frac{5x}{\sqrt{25x^2+1}}[/math] so you forgot a 5.

 

[MacLaddy1]

Ah, of course. I tried that problem repeatedly and continued making the same mistake.

Thanks again, Jameson.