# Indefinite Integral - II

1. Apr 28, 2013

### Saitama

1. The problem statement, all variables and given/known data
$$\int \left(\frac{x}{x\cos x-\sin x} \right)^2 dx$$

2. Relevant equations

3. The attempt at a solution
Factoring out $\cos x$ from the denominator, the integral transforms to
$$\int \sec^2x \left(\frac{x}{x-\tan x}\right)^2dx$$
Substituting $\tan x=t$, $\sec^2 xdx=dt$
$$\int \left(\frac{\arctan t}{\arctan t-t}\right)^2dt$$
I honestly have no idea about how to proceed from here.

Any help is appreciated. Thanks!

2. Apr 28, 2013

### SteamKing

Staff Emeritus
I don't think you can factor cos x from sin x and obtain tan x. I suggest you expand the terms within the brackets and see if it can be simplified with trig identities.

3. Apr 28, 2013

### Saitama

Why can't I factor out cosx?
sinx/cosx=tanx

4. Apr 28, 2013

### SteamKing

Staff Emeritus
The problem is you have only sin x in your integral.

5. Apr 28, 2013

### haruspex

Look at the derivative of x cos(x) - sin(x). This suggests a way to use integration by parts.

6. Apr 28, 2013

### Saitama

The derivative of xcos(x)-sin(x) is -xsin(x) but I still can't figure it out. The whole thing is squared, I am completely lost.

7. Apr 28, 2013

### SteamKing

Staff Emeritus
You could have expanded the brackets by now and checked to see if trig identities could further simplify the integrand.

8. Apr 28, 2013

### haruspex

It suggested to me writing the integrand as $\frac{x}{\sin x} \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}$. Try integrating that by parts.

9. Apr 29, 2013

### Saitama

Integrating by parts:
$$\frac{x}{\sin x} \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx-\int \frac{\sin x-x\cos x}{\sin^2 x} \left(\int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx \right)dx$$

$$\because \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx=(x\cos x-\sin x)$$
The integral simplifies to:
$$\frac{x}{\sin x}(x\cos x-\sin x)+\int \left(\frac{x \cos x-\sin x}{\sin x} \right)^2dx$$
If I expand the squared term, I get an integral involving cot and cot^2, should I integrate by parts again?

10. Apr 29, 2013

### haruspex

Retry that step.

11. Apr 29, 2013

### SammyS

Staff Emeritus
[STRIKE]How is that integration by parts? What were the parts you used?[/STRIKE]

Try $\displaystyle \ u=\frac{x}{\sin(x)}\ \text{ and }\ dv=\frac{x\sin(x)}{\left(x\cos(x)-\sin(x) \right)^2}\,dx \ .$

Added in Edit: Part of the above edited out.
I see it now .

Last edited: Apr 29, 2013
12. Apr 29, 2013

### haruspex

Looks to me that Pranav-Arora was getting there - just made a mistake in the first integration.

13. Apr 29, 2013

### SammyS

Staff Emeritus
Oh ! I see it now.

I had looked at it several times and didn't see how that fit integration by parts.

14. Apr 30, 2013

### Saitama

Oops, its the reciprocal of that.

This time I get:
$$\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}+\int \csc^2 xdx=\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}-\cot x$$

But on further simplification, I do not end up with the right answer.

15. Apr 30, 2013

### ehild

It is correct, bring to common denominator and simplify with sinx.Show what you did.

ehild

16. Apr 30, 2013

### Saitama

Got it, thanks!

Writing cot in terms cos and sin
$$\frac{x-x\cos^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x(1-\cos^2 x)+\cos x \sin x}{\sin x(x\cos x-\sin x)}$$
As 1-cos^2x=sin^2x
$$\frac{x\sin^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x\sin x+\cos x }{x\cos x-\sin x}$$

Thanks a lot haruspex!

Last edited: Apr 30, 2013