1. ## Integral of cos(x)/sin^2(x).

Hello guys here is my problem uploaded: , see the attached image of my scanned paper.

2. Originally Posted by Riazy
Hello guys here is my problem uploaded:
Your second attempt is almost correct (you're missing a minus sign on the final answer...and its csc x, not cot x).

1) Make the substitution $\displaystyle u=\sin x$. Then $\displaystyle \displaystyle\int\dfrac{\cos x}{\sin^2x}\,dx\xrightarrow{u=\sin x}{}\int\dfrac{\,du}{u^2}=\ldots$

2) Observe that $\displaystyle \dfrac{\cos x}{\sin^2 x}=\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}=\csc x\cot x$ and recall that $\displaystyle \dfrac{\,d}{\,dx}\left[\csc x\right]=-\csc x\cot x$.

So either way you do this, you should end up with $\displaystyle \displaystyle\int\dfrac{\cos x}{\sin^2x}\,dx=-\csc x+C$.

3. Your substitution should be $\displaystyle \displaystyle u = \sin{x}$, not $\displaystyle \displaystyle u = \sin^2{x}$. Then the integral reduces to

$\displaystyle \displaystyle \int{u^{-2}\,du}$.

4. I have a problem though, My key answer book gives the answer -1/sinx + c. I am a bit confused now, are there different answers?

5. You should know that $\displaystyle \displaystyle \frac{1}{\sin{x}} = \csc{x}$...

6. Oh, over here in sweden we don't use csc x, not atleast in my bachelor civil engineering program, Thanks alot BTW! both of u

7. I would have done this slightly differently.

It has been proven that for any function f(x), $\displaystyle \displaystyle \int f'(x)f(x)^n dx=\frac{1}{n+1}[f(x)^{n+1}]+c$

This applies here if you write this as:

$\displaystyle [Cos(x)][Sin(x)^{-2}]$

as f(x) = sin x; n = -2; f'(x) = Cos(x)

Substituting into:

$\displaystyle \frac{1}{n+1}[f(x)^{n+1}]+c$

Gives $\displaystyle \frac{1}{-2+1}[Sin(x)^{-2+1}]+c$

$\displaystyle =\frac{1}{-1}[Sin(x)^{-1}]+c$

$\displaystyle =\displaystyle\frac{-1}{Sin(x)}$

$\displaystyle =-\csc(x) + c$

8. Hello, Riazy!

I did it like this . . .

$\displaystyle \displaystyle \int \frac{\cos x}{\sin^2\!x}\,dx \;=\;\int\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\,dx \;=\;\int \csc x \cot x\,dx \;=\;-\csc x + C$

9. Originally Posted by Soroban
Hello, Riazy!

I did it like this . . .

$\displaystyle \displaystyle \int \frac{\cos x}{\sin^2\!x}\,dx \;=\;\int\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\,dx \;=\;\int \csc x \cot x\,dx \;=\;-\csc x + C$

I don't know how you can all remember integrals like $\displaystyle \displaystyle \int{\csc{x}\cot{x}\,dx}$ or $\displaystyle \displaystyle \int{f'(x)[f(x)]^n\,dx} = \frac{1}{n+1}[f(x)]^{n+1} + C$ off the top of your head. I prefer to remember the procedures, and in this case, to reduce everything to sines and cosines and use substitution - at least then I only have to remember a couple of fundamental integrals

10. Originally Posted by Prove It
I don't know how you can all remember integrals like $\displaystyle \displaystyle \int{\csc{x}\cot{x}\,dx}$ or $\displaystyle \displaystyle \int{f'(x)\,f^n(x)\,dx} = \frac{1}{n+1}f^{n+1}(x) + C$ off the top of your head. I prefer to remember the procedures, and in this case, to reduce everything to sines and cosines and use substitution - at least then I only have to remember a couple of integrals
I don't always remember the integrals, I often just remember how to derive them from differentiation.

11. You should be able to remember"
If, in the integral $\displaystyle \int sin^n(x)cos^m(x)dx$, n is odd, say n= 2k+ 1, you can factor out one "sine" to use with the differential: $\displaystyle \int sin^{2k}(x) cos^m(x) (sin(x)dx)$, then use $\displaystyle sin^2(x)= 1- cos^2(x)$ to get
$\displaystyle \int (1- cos^2(x))^k cos^m(x)(sin(x)dx)$
and now the substitution u= cos(x), so that du= -sin(x)dx, makes that a polynomial integtration:
$\displaystyle -\int (1- u^2)^k u^m u$
Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.

(If both m and n are even, you can use $\displaystyle sin^2(x)= (1/2)(1- cos(2x)$ and/or $\displaystyle cos^2(x)= (1/2)(1+ cos(2x))$ to reduce the powers.)

12. Originally Posted by HallsofIvy
You should be able to remember"

Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.

(If both m and n are even, you can use $\displaystyle sin^2(x)= (1/2)(1- cos(2x)$ and/or $\displaystyle cos^2(x)= (1/2)(1+ cos(2x))$ to reduce the powers.)
Oh! Is that ALL? Well, thanks for clearing that up! That's so much simpler and easier to remember!

13. Originally Posted by HallsofIvy
You should be able to remember"

Of course, if m is odd, you can do the same thing, switching "sine" and "cosine".

This also works if either m or n (or both) are negative but gives a "rational function" integtral rather than a polynomial integral.

The specific formulas others gave may be quicker for specific values of m or n, but that is, in my opinion, the simplest general method.

(If both m and n are even, you can use $\displaystyle sin^2(x)= (1/2)(1- cos(2x)$ and/or $\displaystyle cos^2(x)= (1/2)(1+ cos(2x))$ to reduce the powers.)
I don't know HOW I could have been so blind... How DID I get by just remembering the integrals of $\displaystyle \displaystyle \sin{x}, \cos{x}$ and $\displaystyle \displaystyle \sec^2{x}$? :P

14. Originally Posted by Prove It
I don't know how you can all remember integrals like $\displaystyle \int{f'(x)[f(x)]^n\,dx} = \frac{1}{n+1}[f(x)]^{n+1} + C$
It's very hard to forget that integral once you've read it 'backwards' (well, sort of):

$\displaystyle \left(\frac{1}{n+1}[f(x)]^{n+1}\right)' = f'(x)[f(x)]^n$

... because it's very a chain rule picturesque. Or I memorise things in weird ways!

15. Originally Posted by TheCoffeeMachine
It's very hard to forget that integral once you've read it 'backwards' (well, sort of):

$\displaystyle \left(\frac{1}{n+1}[f(x)]^{n+1}\right)' = f'(x)[f(x)]^n$

... because it's very a chain rule picturesque. Or I memorise things in weird ways!
I do something similar to learn it, so you're only about as crazy as I am. Which is probably not very reassuring.

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