a problem about integral

**mami** · January 1st 2012, 11:23 AM

could you help me please. any help will be appreciated...

**Siron** · January 1st 2012, 12:56 PM

Have you tried with induction? (Maybe that will work)

**FernandoRevilla** · January 2nd 2012, 08:28 AM

Originally Posted by mami

could you help me please. any help will be appreciated...

Could you help us mentioning the subject where that integral is included? For example, sometimes we solve an integral using Complex Variable and we receive the following answer: Sorry, we have not covered Complex Variable, etc., etc., etc.

**MarkFL** · January 2nd 2012, 07:58 PM

I have become interested in this problem and have spent a great deal of time trying to get induction (among other things) to work, but have not been successful. Hopefully, someone can show me where I am going wrong. Obviously the statement is true for the base case of n = 1. So I stated the induction hypothesis $P_n$ :

$\int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin(na)}{\sin(a)}$

Assuming this is true, I then added $P_{n+2}$ to get:

$\int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx+\int_0^{\pi}\frac{\cos\left((n+2)x \right)-\cos\left((n+2)a \right)}{\cos(x)-\cos(a)}\,dx=$

$\pi\frac{\sin(na)}{\sin(a)}+\pi\frac{\sin\left((n+ 2)a\right)}{\sin(a)}$

$\int_0^{\pi}\frac{\left(\cos\left((n+2)x \right)+\cos(nx)\right)-\left(\cos\left((n+2)a\right)+\cos(na) \right)}{\cos(x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}\left(\sin\left((n +2)a\right)+\sin(na)\right)$

$\int_0^{\pi}\frac{\left(2\cos\left((n+1)x \right)\cos(x)\right)-\left(2\cos\left((n+1)a\right)\cos(a)\right)}{\cos (x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}$2\sin\left((n+1) a\right)\cos(a)$$

$\int_0^{\pi}\frac{\cos\left((n+1)x\right) \frac{\cos(x)}{\cos(a)}-\cos\left((n+1)a\right)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin\left((n+1)a \right)}{\sin(a)}$

I can't figure out how to get rid of the $\frac{\cos(x)}{\cos(a)}$ in the numerator of the integrand.

edit: sorry about my edits...this board has some unexpected idiosyncrasies concerning the
\left and \right operators in $\LaTeX$ ...

**MarkFL** · January 2nd 2012, 08:33 PM

I should add that this problem has also been posted on the forum which I help moderate, and I have posted the same problem I have in using induction.

The OP is an infrequent visitor there, and it may be a while before we hear from them to get any feedback on the context of the problem.

**MarkFL** · January 4th 2012, 09:29 AM

The following fix was given to me:

$\int_0^{\pi}\frac{\cos((n+1)x)}{\cos(a)}\,dx=0$

$\int_0^{\pi}\frac{\cos((n+1)x)\left(\frac{\cos(x)} {\cos(a)}-1\right)+\cos((n+1)a)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=0$

$\int_0^{\pi}\frac{\cos((n+1)x)\frac{\cos(x)}{\cos( a)}-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=\int_0^{\pi}\frac{\cos((n+1)x)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx$

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