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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

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.
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