# Another integral

1. Jul 26, 2004

### Ornum

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.

2. Jul 26, 2004

### pnaj

$${d \over {dx}}({1 \over {f(x)}}) = {{ - f'(x)} \over {f(x)^2 }}$$

3. Jul 26, 2004

### AKG

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:

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

4. Jul 26, 2004

### Ornum

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.

5. Jul 27, 2004

### pnaj

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.

6. Jul 27, 2004

### pnaj

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.

7. Aug 7, 2004

### Dr.Brain

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. Aug 7, 2004

### maverick280857

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 ;-)