How to Solve the Integral of cot(x)/sin(x)?

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SUMMARY

The integral of cot(x)/sin(x) can be solved by rewriting it as the integral of cos(x)/sin²(x). By substituting u = sin(x), the integral simplifies to ∫(du/u²), resulting in -csc(x) + C. Additionally, integration by parts can be applied using the ILATE rule, which prioritizes inverse trigonometric functions, logarithmic functions, algebraic functions, trigonometric functions, and exponential functions for optimal results.

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Ornum
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Hello again people, thank you for the help I received 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:

\int \frac{\cot x}{\sin x}\,dx

If anyone can aid me in solving this I would be very glad. Thanks in advance.
 
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This might help you to see it:

<br /> {d \over {dx}}({1 \over {f(x)}}) = {{ - f&#039;(x)} \over {f(x)^2 }}<br />
 
Ornum said:
Hello again people, thank you for the help I received 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:

\int \frac{\cot x}{\sin x}\,dx

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:

\int \frac{\cos x}{\sin ^2 x}\,dx

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

\int \frac{du}{u^2} = -u^{-1} + C = -\csc x + C
 
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:
 
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:
 
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:
 
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...
 
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 \int u dv.

<br /> \int udv = uv - \int vdu<br />

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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