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Integerating (1/((sinx)^2))

  • Thread starter Geekchick
  • Start date
1. Homework Statement

25[tex]\int\frac{1}{sin^{2}x}[/tex]dx


3. The Attempt at a Solution

I wasn't sure if I could change [tex]\frac{1}{sin^{2}x}[/tex] to Csc[tex]^{2}[/tex]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...
 
FYI, the original problem was [tex]\int\frac{1}{x^{2}\sqrt{25-x^{2}}}[/tex]dx I used trigonometric substitution to get to the problem above.
 
192
0
Well in an online integral table I found that [tex]\int \csc^2 ax dx = -\frac{1}{a} \cot ax + C[/tex] so you should probably come up with [tex]-\frac{cot(25x)}{25} + C[/tex].
 
316
0
-25Cot(x) is the integral of 25/sin^2(x), you can check it by computing the derivative.
 
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.
 
Oh I did catch that it should be 1/25 not 25. But its still slightly off.
 

Dick

Science Advisor
Homework Helper
26,258
618
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.
 
I substituted [tex]\frac{\sqrt{25-x^{2}}}{x}[/tex] for cot So what I end up with is -[tex]\frac{\sqrt{25-x^{2}}}{25x}[/tex]+c
 

Dick

Science Advisor
Homework Helper
26,258
618
That looks fine to me.
 

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