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Let $ninmathbb{N}$ and $xin]0,pi[$, I am asked to calculate the following :
$$ I_n = int_0^{pi} dfrac{cos(nx)cos(x) - cos(nt)cos(t)}{cos(x) -cos(t)}dt$$
From testing on small values of $n$, it seems that this integral is equal to $npicdot cos^n(x)$ but I can't seem to prove it. I tried finding a recurrence formula but didn't succeed.

Here is my working for $n=0$, $n=1$ and $n=2$ :
For $n=0$,
$$ I_0=int_0^{pi}dfrac{cos(x) -cos(t)}{cos(x) - cos(t)}dt = pi$$
For $n=1$,
$$ I_1 = int_0^{pi} dfrac{cos^2(x) -cos^2(t)}{cos(x)-cos(t)}dt=int_0^{pi}cos(x) + sin(t)dt = picdot cos(x)$$
For $n=2$ :
$$ I_2 = int_0^{pi} dfrac{2cos^3(x) - 2cos^3(t) -cos(x) + cos(t)}{cos(x) - cos(t)}dt$$
$$ I_2 = 2int_0^{pi}cos^2(x) +cos(x)cos(t) + cos^2(t) dt - pi$$
$$ I_2 = 2picos^2(x) + int_0^{pi}cos(2t)+1dt - pi$$
$$ I_2 = 2picos^2(x) $$

This is my first post here, please tell me if I did anything wrong. I tried searching this integral on this website without any success.

Here's a solution that only rests on the following simple trigonometric identity:
$$cos(a+b)+cos(a-b)=2cos(a)cos(b)tag{1}$$
We'll get back to it later, but for now, notice that
$$begin{split}
I_n(x)&=int_0^{pi} frac{cos(nx)cos(x) - cos(nt)cos(t)}{cos(x) -cos(t)}dt\
&=int_0^{pi}frac{[cos(nx)-cos(nt)]cos(x) + cos(nt)[cos(x)-cos(t)]}{cos(x) -cos(t)}dt\
&=cos(x)int_0^{pi}frac{cos(nx)-cos(nt)}{cos(x) -cos(t)}dt+int_0^picos(nt)dt
end{split}$$
In other words,
$$I_n(x)=cos(x)J_n(x)+pidelta_{n=0}tag{2}$$
where we define $$J_n(x)=int_0^pi frac{cos(nx)-cos(nt)}{cos(x)-cos(t)}dt$$
and the Kronecker symbol $delta_{n=0}$, which is equal $0$, unless $n=0$, in which case it's equal to $1$.

Now, let's go back to (1). Plugging $a=nx$ and $b=x$ into that identity implies that
$$cos((n+1)x)+cos((n-1)x)=2cos x cos(nx)$$
Subtracting the same equation with $t$ to this one yields
$$
begin{split}
cos((n+1)x)-cos((n+1)t) \
+cos((n-1)x)-cos((n-1)t)=\
2cos x cos(nx)-2cos(t)cos(nt)
end{split}$$
Dividing by $cos(x)-cos(t)$, and integrating over $[0,pi]$ leads to
$$J_{n+1}(x)+J_{n-1}(x)=2I_n(x)tag{3}$$
Finally, combining [2] and [3] gets us, for $ngeq 0$,
$$J_{n+2}(x)-2cos(x)J_{n+1}(x)+J_{n}(x)=0$$

The solution to this second-order recurrence relation is
$$J_n(x)=alpha e^{inx}+beta e^{-inx}$$
Since, $J_0=0$ and $J_1=pi$,
$$J_n(x)=frac {pi sin(nx)}{sin x}$$
and $$I_n(x)=picos(x)frac{sin(nx)}{sin(x)} mbox{ for } ngeq 1 mbox{, and }I_0=pi$$

A nice way to evaluate generalized integrals is to consider them as the coefficients of an infinite series. Therefore, the coefficient of the $n$th term is simply the integral under question. Before we begin though, there is one identity to note:

$$sumlimits_{ngeq0}z^ncos nx=frac {1-zcos x}{z^2-2zcos x+1}$$
Proof: Rewrite $cos nx$ as the real part of $e^{nix}$. Using the infinite geometric sequences, we get that$$sumlimits_{ngeq0}left(ze^{ix}right)^n=frac 1{1-ze^{ix}}$$Now, take the real part of both sides. Clearly, the left - hand side becomes $z^ncos nx$. Meanwhile, the right - hand side becomes, through some clever rationalization$$begin{align*}operatorname{Re}left[frac 1{1-ze^{ix}}right] & =operatorname{Re}left[frac 1{1-zcos x-zisin x}right]\ & =operatorname{Re}left[frac {1-zcos x+zisin x}{(1-zcos x)^2+z^2sin^2x}right]\ & =frac {1-zcos x}{z^2-2zcos x+1}end{align*}$$completing the proof.

With that in mind, we are ready to begin. Since the OP has stated in the comments that he is trying to evaluate the integral$$I_n=intlimits_0^{pi}mathrm dt,frac {cos nx-cos nt}{cos x-cos t}$$I will show a way to evaluate it in this answer. To wit, denote the generating function of the integral as $G(z)$

$$G(z)=sumlimits_{ngeq0}I_nz^n$$

And remember that the coefficient of $z^n$ simply gives $I_n$. Interchange the sum and the integral, and using the identity we've derived above, get

$$begin{align*}G(z) & =intlimits_0^{pi}frac {mathrm dt}{cos x-cos t}sumlimits_{ngeq0}z^nbiggr[cos nx-cos ntbiggr]\ & =intlimits_0^{pi}frac {mathrm dt}{cos x-cos t}left[frac {1-zcos x}{z^2-2zcos x+1}-frac {1-zcos t}{z^2-2zcos t+1}right]end{align*}$$

Combining the two fractions, and recalling that any terms in $z$ are constants, the function becomes

$$G(z)=frac {z(1-z^2)}{z^2-2zcos x+1}intlimits_0^{pi}frac {mathrm dt}{z^2-2zcos t+1}$$

The remaining integral can be easily evaluated using a Weierstrass substitution. Substitute $w=tanleft(tfrac t2right)$ so that

$$begin{array}{|c|c|c|}hline w=tanleft(dfrac t2right) & mathrm dt=dfrac {2,mathrm dw}{1+w^2} & cos t=dfrac {1-w^2}{1+w^2}\hlineend{array}$$

The remaining rational function can be evaluated in an elementary fashion

$$begin{align*}G(z) & =frac {2z(1-z^2)}{z^2-2zcos x+1}intlimits_0^{infty}frac {mathrm dw}{w^2(1+z)^2+(1-z)^2}\ & =frac {2z}{z^2-2zcos x+1}arctanleft(frac {1+z}{1-z}wright),Biggrrvert_0^{infty}\ & =frac {pi z}{z^2-2zcos x+1}end{align*}$$

From the second line, it's important to observe that the argument of the inverse tangent will remain positive if and only if $|z|<1$. When $|z|<1$, then the denominator is positive, as well as the numerator. Therefore, $tfrac {1+z}{1-z}>0$. However, if $|z|>1$, then the argument is less than zero and there is an extra negative sign. For the purpose of this question, we'll consider when $|z|<1$.

Now all we have to do is find the coefficient of $z^n$. There is a nice and convenient way to do this by using

$$2cos x=e^{ix}+e^{-ix}$$

Factoring the denominator by grouping gives

$$begin{align*}frac z{z^2-2zcos x+1} & =frac z{(1-ze^{ix})(1-ze^{-ix})}\ & =zsumlimits_{kgeq0}z^k e^{kix}sumlimits_{lgeq0}z^l e^{-lix}end{align*}$$

Now observe what happens when we expand the products together$$begin{multline}(1+ze^{ix}+z^2e^{2ix}+cdots)(1+ze^{-ix}+z^2e^{-ix}+cdots)\=1+z(e^{ix}+e^{-ix})+z^2(e^{2ix}+1+e^{-2ix})+cdotsend{multline}$$

The sum within the parenthesis seems to start off at the index of the $n$th term and decrease by a factor of two! Using this, it's possible to rewrite the coefficients conveniently as

$$a_k=sumlimits_{m=0}^ke^{(k-2m)ix}=frac {sin x(k+1)}{sin x}$$

Hence$$frac {pi z}{z^2-2zcos x+1}=pisumlimits_{kgeq1}frac {sin xk}{sin x}z^k$$

And setting $k$ as $n$ gives the term $z^n$. Therefore, our integral is simply$$intlimits_0^{pi}mathrm dt,frac {cos nx-cos nt}{cos x-cos t}color{blue}{=frac {pisin xn}{sin x}}$$

Completing Frank's solution:

$$ [z^n]frac{pi z}{z^2-2zcos x+1} = frac{pi}{2}[z^{n}]left(frac{1}{z-e^{ix}}+frac{1}{z-e^{-ix}}right) $$
equals, by geometric series,
$$ frac{pi}{2}left(-e^{-(n+1)ix}-e^{(n+1)ix}right)=-picos((n+1)x). $$

I will do the integral for $n=3$, unfortunately this will disprove your given result. (unless I did a mistake)
$$I_3= int_0^pi frac{cos(3x)cos x -cos(3t)cos t}{cos x-cos t}dt$$
Since $cos(3 y)=4cos^3 y -3cos y$ we have: $$cos(3x)cos x -cos (3t)cos t=4(cos^4 x-cos^4t)-3(cos^2 x-cos^2 t)$$
$$=4(cos x-cos t)(cos x+cos t) (cos^2 x+cos^2t)-3(cos x-cos t)(cos x+cos t)$$
$$Rightarrow I_3=int_0^pi (cos x+cos t)(4(cos ^2 x+cos^2 t)-3)dt$$
$$overset{pi-tto t}=int_0^pi (cos x-cos t)(4(cos^2 x+cos^2 t)-3)dt$$
$$Rightarrow 2I_3=2cos xint_0^pi (4(cos^2 x+cos^2 t)-3)$$
$$Rightarrow I_3= 4pi cos^3 x +cos x underbrace{int_0^pi cos^2 tdt}_{=frac{pi}{2}}-3pi cos x=2pi(cos x+2cos(3x))$$

For $n=4$ we have: $$cos(4x)=8cos^4 x-8cos^2 x+1$$
Denoting $cx=cos x$ and $ct=cos t,$ we get the integrand to be:
$$frac{8(cx^5-ct^5)-8(cx^3-ct^3)+(cx-ct)}{cx-ct}$$
And using the fact that: $$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4$$
$$Rightarrow I_4=8int_0^pi(cx^4+cx^3 ct+cx^2 ct^2 +cx ct^3+ct^4)dt$$
$$-8int_0^pi (cx^2+cxct+ct^2) dt +int_0^pi dt$$
We have: $$int_0^pi ct dt=int_0^pi ct^3 dt =0$$
$$int_0^pi ct^2 dt= frac{pi}{2}, int_0^pi ct^4 dt=frac{3pi}{8}$$
$$Rightarrow I_4=(8pi cx^4 +4pi cx^2 +3pi )-(8pi cx^2 +4pi)+pi$$
$$=8pi cx^4 -4pi cx^2 =4pi cos^2 x cos(2x)$$