1. ## a problem about integral

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

2. ## Re: a problem about integral

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

3. ## Re: a problem about integral

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.

4. ## Re: a problem about integral

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 $\displaystyle P_n$:

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

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

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

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

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

$\displaystyle \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)}$$2\sin\left((n+1) a\right)\cos(a)$$$

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

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

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

5. ## Re: a problem about integral

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.

6. ## Re: a problem about integral

The following fix was given to me:

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

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

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