View Full Version : Another integral
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:
\int \frac{\cot x}{\sin x}\,dx
If anyone can aid me in solving this I would be very glad. Thanks in advance.
This might help you to see it:
{d \over {dx}}({1 \over {f(x)}}) = {{ - f'(x)} \over {f(x)^2 }}
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:
\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:
Dr.Brain
Aug7-04, 06:27 AM
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...
maverick280857
Aug7-04, 09:40 AM
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.
\int udv = uv - \int vdu
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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