If $\displaystyle \int^{\pi}_{0}\ln(1-2a\cos x+a^2)dx=k\ln(a),a>0$
Try: assume $$I(a)=\int^{\pi}_{0}\ln(1-2a\cos x+a^2)dx=k\ln(a)$$
Differentiate both side with respect to $a$
$$I'(a)=\int^{\pi}_{0}\frac{2a-2\cos x}{1-2a\cos x+a^2}dx=\frac{k}{a}$$
Now substitute $a=1$ because given $a>0$
So $$\int^{\pi}_{0}\frac{2-2\cos x}{2-2\cos x}dx=k.$$
So $k=\pi.$
But answer given is $2\pi$
Could some help me what is wrong in my process, thanks
You should instead have ( wrong differentiation ) $$ I'\left(a\right)=\int_{0}^{\pi}\frac{2a-2\cos\left(x\right)}{1-2a\cos\left(x\right)+a^2}\text{d}x $$
Edit : You cannot evaluate in $1$ because the function is only diffentiable on $\left]1,+\infty\right[$ ( or $\left[0,1\right[ ). This subject was marked as duplicate but I dont think it answered why OP was wrong.
It's because you differentiated under the integral sign without proving you can. If you consider $$ f\left(a,x\right)=\ln\left(1-2a\cos\left(x\right)+a^2\right) $$ for $\left(a,x\right) \in \mathbb{R}^{*+} \times \left]0,\pi\right[$ then $$ \frac{\partial f}{\partial a}\left(a,x\right)=\frac{2a-2\cos\left(x\right)}{1-2a\cos\left(x\right)+a^2} $$ To show you can differentiate under the integral sign you need to find $\displaystyle \varphi \in \ell^{1}\left(\left]0,\pi\right[\right)$ such as $$ \left|\frac{\partial f}{\partial a}\left(a,x\right)\right| \leq \varphi\left(x\right) $$ You see that you cannot makes $a$ disappear unless you consider a compact set.
I dont have time I must go, however what is import to notice is that for $a=1$, $$ 1-2a\cos(x)+a^2=2-2\cos(x) $$ And it can be equal to $0$ ( when $x=0$ and $0 \in \left[0, \pi\right]$ ! ) . Intuitively, you couldnt simplify it to $1$ when evaluating to $a=1$ because you cannot divide by $0$ and it could have taken the value $0$. However, try to choose $a=2$ you'll find $k=2\pi$.
