1. ## trig substitutions

I have the integration of radical(25 - x^2)/(x)
i set x = 5sinx
dx = 5cosx
i ended up with integral of cos^2(x)/sin(x)
..not so sure where to go from here. Thanks for the help.

2. Originally Posted by davecs77
I have the integration of radical(25 - x^2)/(x)
i set x = 5sinx
dx = 5cosx
i ended up with integral of cos^2(x)/sin(x)
..not so sure where to go from here. Thanks for the help.
$\int \frac{\sqrt{25 - x^2}}{x} dx$

Let $x = 5 sin(u)$ --> $dx = 5 cos(u) du$

Thus
$\int \frac{\sqrt{25 - x^2}}{x} dx = \int \frac{\sqrt{25 - 25sin^2(u)}}{5sin(u)} 5 cos(u) du$

$= 5 \sqrt{5} \int \frac{cos^2(u)}{sin(u)}du$

$= 5 \sqrt{5} \int \frac{cos^2(u) - 1 + 1}{sin(u)}du$

$= 5 \sqrt{5} \int \frac{cos^2(u) - 1}{sin(\theta)} du + 5 \sqrt{5} \int \frac{1}{sin(u)}du$

$= -5 \sqrt{5} \int sin(\theta) du + 5 \sqrt{5} \int \frac{1}{sin(u)}du$

Is that any better?

-Dan

3. Originally Posted by topsquark
$\int \frac{\sqrt{25 - x^2}}{x} dx$

Let $x = 5 sin(u)$ --> $dx = 5 cos(u) du$

Thus
$\int \frac{\sqrt{25 - x^2}}{x} dx = \int \frac{\sqrt{25 - 25sin^2(u)}}{\sqrt{5} \sqrt{sin(u)}} 5 cos(u) du$

$= 5 \sqrt{5} \int \frac{cos^2(u)}{\sqrt{sin(u)}}du$

Can you take it from here?

-Dan
I dont really see what you can do with the intergral of cos^2(u)/rad(sin(u)
the 2nd power in the cos makes it tough.

4. Originally Posted by davecs77
I dont really see what you can do with the intergral of cos^2(u)/rad(sin(u)
the 2nd power in the cos makes it tough.
Ugh. I must have been tired last night. I have fixed my previous post.

-Dan