Integrating Inverse Trigonometric Functions: How to Solve for 1/(sin^2x)?

[Geekchick]

Homework Statement

25\int\frac{1}{sin^{2}x}dx

The Attempt at a Solution

I wasn't sure if I could change \frac{1}{sin^{2}x} to Csc^{2}x but when I did I ended up with -25Cotx which when I checked the integral in my calculator and it was wrong. So now I'm lost...

 

[Geekchick]

FYI, the original problem was \int\frac{1}{x^{2}\sqrt{25-x^{2}}}dx I used trigonometric substitution to get to the problem above.

 

[kbaumen]

Well in an online integral table I found that \int \csc^2 ax dx = -\frac{1}{a} \cot ax + C so you should probably come up with -\frac{cot(25x)}{25} + C.

 

[yyat]

-25Cot(x) is the integral of 25/sin^2(x), you can check it by computing the derivative.

 

[Geekchick]

Well did I go wrong before I got to the sin integral? because when I checked it against the original problem it didn't match.

 

[Geekchick]

Oh I did catch that it should be 1/25 not 25. But its still slightly off.

 

[Dick]

Judging from what you've shown us, you used the substitution x=5*sin(u) to reduce the integral to (1/25) times the integral (1/sIn(u)^2)*du. That's fine. So you've got -cot(u)/25 as the integral. You still have to express that in terms of x.

 

[Geekchick]

I substituted \frac{\sqrt{25-x^{2}}}{x} for cot So what I end up with is -\frac{\sqrt{25-x^{2}}}{25x}+c

 

[Dick]

That looks fine to me.