# Thread: Find the Indefinite Integral (cosx/1-cos^2x)dx

1. ## Find the Indefinite Integral (cosx/1-cos^2x)dx

Find the indefinite Integral (cosx/1-cos^2x)dx.

Thanks so much for your help in avance!

2. Originally Posted by Safety33
Find the indefinite Integral (cosx/1-cos^2x)dx.

Thanks so much for your help in avance!
$1-\cos^2(x)=\sin^2(x)$

So we have

$\int\frac{\cos(x)}{\sin^2(x)}dx$

Now let $u=\sin(x)\Rightarrow{du=\cos(x)}$

So now we have

$\int\frac{du}{u^2}$

You can go from there, yeah?

3. $\int \frac{\cos x}{1-\cos^2 x}~dx$

The most known trigonometric identity: $\sin^2 x + \cos^2 x = 1$.

So $1-\cos^2 x = \sin^2 x$.

$\int \frac{\cos x}{\sin^2 x}~dx$

Substitute $u=\sin x$...

4. This is actually a very easy one.

Rewrite as:

$\int cot(x)csc(x)dx$

Now, let $u=csc(x), \;\ du=-cot(x)csc(x)dx$

Now you get:

$-\int du$

$=-u$

$=-csc(x)$

Add a constant C if you need to.

5. Im coming out with 1/sinx. Is that incorrect?

I have U^-1, so 1/U=1/sinx

6. Originally Posted by Safety33
Im coming out with 1/sinx. Is that incorrect?

I have U^-1, so 1/U=1/sinx
$\int\frac{du}{u^2}=\frac{{\color{red}{-1}}}{u}$

7. Where does the -1 come from? Im sorry Im struggling so much with this.

8. Originally Posted by Safety33
Where does the -1 come from? Im sorry Im struggling so much with this.
$\int{u^n}du=\frac{u^{n+1}}{n+1}\quad\text{provided that }n\ne{-1}$

So we have

$\int\frac{du}{u^2}=\int{u^{-2}du}=\frac{u^{-1}}{-1}=-u^{-1}=\frac{-1}{u}$

9. Im such and idiot! Thank you very much for walking me through that.

10. Originally Posted by Safety33
Im such and idiot! Thank you very much for walking me through that.

11. $\frac{-1}{sin(x)}=-csc(x)$.

Same thing. You are good.

12. Originally Posted by Mathstud28

$\int\frac{du}{u^2}=\frac{{\color{red}{-1}}}{u}+{\color{blue}k}$

13. Originally Posted by Krizalid
$\frac{-1}{u}+{\color{red}\bold{C}}$

...who does k anyways..

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