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.