# Stuck on this integral

1. Jan 17, 2005

### Physicsisfun2005

integral of 1/(cscx-1)

i can't think of any trig subs or anything......natural log prob? ;)....i udnno i'm truely clueless. all i need is a hint or more if u like ;)

also, am i using integrals for this prob:

Compute the average value of f(x)=x/(x+3) over the interval [-a,a], 0<a<3

I think i will just get an expression for the answer

Last edited: Jan 17, 2005
2. Jan 17, 2005

### Benny

Here is what I have done. So you should go through my working to check for errors.

$$\int {\frac{1}{{\cos ecx - 1}}} dx$$

$$= \int {\frac{1}{{\left( {\frac{1}{{\sin x}} - 1} \right)}}} dx$$

$$= \int {\frac{1}{{\left( {\frac{{1 - \sin x}}{{\sin x}}} \right)}}} dx$$

$$= \int {\frac{{\sin x}}{{1 - \sin x}}dx}$$

When I see something like the above I just think, something needs to be done to the denominator. I would use a technique similar to rationalising a surd expression as follows.

$$= \int {\frac{{\sin x}}{{1 - \sin x}} \times \frac{{1 + \sin x}}{{1 + \sin x}}dx}$$

$$= \int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}dx}$$

Now use some trig identities as follows.

$$= \int {\frac{{\sin x + \frac{1}{2} - \frac{1}{2}\cos 2x}}{{\cos ^2 x}}} dx$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{\cos 2x}}{{\cos ^2 x}}dx}$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{2\cos ^2 x - 1}}{{\cos ^2 x}}} dx$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{2\cos ^2 x}}{{\cos ^2 x}}} dx - \frac{1}{2}\int {\frac{{\left( { - 1} \right)}}{{\cos ^2 x}}dx}$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \int {dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} }$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\frac{1}{{\cos ^2 x}}} - \int {dx} }$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\sec ^2 x} dx - \int {dx} }$$

The first integral in the above line can be done by substitution or simply by inspection which can save time.

$$= \frac{1}{{\cos x}} + \tan x - x + c$$

$$= \sec x + \tan x - x + c$$

Last edited: Jan 18, 2005
3. Jan 18, 2005

### dextercioby

Yes,Benny,good work... Though you could have shortcutted the part with the double angle on the cosine...

Daniel.

4. Jan 18, 2005

### Benny

Thanks dextercioby.

I see what you mean, I went around in circles. The question could have been finished as follows.

$$\int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}} dx$$

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

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\tan ^2 x} dx}$$

$$= \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\left( {\sec ^2 x - 1} \right)} dx}$$ The integral on the left can be integrated by substitution or using the fact that $$\frac{d}{{dx}}\left( {\sec x} \right) = \tan x\sec x$$

So $$\int {\frac{1}{{\cos ecx - 1}}dx} = \sec x + \tan x - x + c$$.

5. Jan 18, 2005

### dextercioby

I was thinking about something else...Right from the beginning...It doesn't really matter.It would have been another (more simple) trick besides the one you pulled...

Daniel.

P.S.These 2 tricks shorten the calculus,but the disadvantage is that u may not see either of them... :tongue2:

6. Jan 18, 2005

### poolwin2001

In the numerator you could have used Sinx=-(-Sinx)=-(1-Sinx-1)
which would have simplified the calc instead of *ing and /ing by 1+Sinx

Last edited: Jan 18, 2005
7. Jan 18, 2005

### dextercioby

OMG,how can u say that?? :surprised Would u agree that what u said is basically an alambicated way of:
$$\sin x=(\sin x-1)+1$$

and the minus would appear naturally (when symplifying the denominator and the numerator),not forced...??

Daniel;

PS.Benny,that's the trick i mentioned...

8. Jan 20, 2005

### poolwin2001

What do you mean by natural and forced??
Yes I was talking about this only.$\sin x=(\sin x-1)+1$
I added the -ve sign to prevent confusion,but it seems that it didn't work.