Another integral

  • Thread starter Ornum
  • Start date
  • #1
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Hello again people, thank you for the help I recieved on the last integral I posted here but never got around to replying to, the help was much appreciated. However i've encountered another integral which for some reason I just cannot seem to solve, being:

[itex]\int \frac{\cot x}{\sin x}\,dx[/itex]

If anyone can aid me in solving this I would be very glad. Thanks in advance.
 

Answers and Replies

  • #2
96
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This might help you to see it:

[tex]
{d \over {dx}}({1 \over {f(x)}}) = {{ - f'(x)} \over {f(x)^2 }}
[/tex]
 
  • #3
AKG
Science Advisor
Homework Helper
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Ornum said:
Hello again people, thank you for the help I recieved on the last integral I posted here but never got around to replying to, the help was much appreciated. However i've encountered another integral which for some reason I just cannot seem to solve, being:

[itex]\int \frac{\cot x}{\sin x}\,dx[/itex]

If anyone can aid me in solving this I would be very glad. Thanks in advance.
In many situations, it's really simple just to express everything in terms of sine and cosine:

[itex]\int \frac{\cos x}{\sin ^2 x}\,dx[/itex]

Let [itex]u = \sin x[/itex], therefore [itex]dx = du/\cos x[/itex]. Making the substitution:

[itex]\int \frac{du}{u^2} = -u^{-1} + C = -\csc x + C[/itex]
 
  • #4
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Thank you very much, now you've just shown me I cannot believe how I missed that, however that tends to be how it always is. :wink:
 
  • #5
96
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I always get frustrated when something was staring me in the face but I couldn't see it. That's why I thought a hint may be of more help that the 'full' solution.

Paul. :wink:
 
  • #6
96
0
I'd always get frustrated when I couldn't see something staring me in the face. That's why I thought a hint may be of more help that the 'full' solution.

Paul. :wink:
 
  • #7
538
2
or u can write the above integrl as

INTEGRAL OF cosx cosec(sqr)x

take cosec(sqr)x as 2nd function and integrate by parts ...gives u answer instantly...
 
  • #8
1,789
4
Some general advice: after you have an integral in a "standard looking form" (not even a standard form, for you wouldn't be reading my advice if it were in a standard form) try integration by parts keeping the ILATE rule in mind

I = inverse trigonometric function
L = logarithmic function
A = algebraic function
T = trigonometric function
E = exponential function

This order gives you an idea of which function to chose as u and which to chose v, when you wish to evaluate the integral [tex]\int u dv[/tex].

[tex]
\int udv = uv - \int vdu
[/tex]

By the way, Dr. Brain are you from India? My guess is that you're in class 11/12. Correct me if I am wrong ;-)
 

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