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Consider this definite integral
$$I=int_0^2pidfracxcos x1 +cos xdx$$

Method 1

$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$

$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$

Adding (1) and (2)

$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$

$$I=pi int_0^2pidfraccos x1 +cos xdx$$

$$I=2pi int_0^pidfraccos x1 +cos xdx.....(P1)$$

$$I=2piint_0^pidfrac1+cos x-11 +cos xdx$$

$$I=2pi int_0^pi1-dfrac11 +cos xdx$$

$$I=2pi int_0^pi1dx-2pi int_0^pidfrac11 +cos xdx$$

$$I=2pi[x]_0^pi -2pi int_0^pidfrac12cos ²dfracx2dx$$

$$I=2pi ² - pi int_0^pi sec ²dfracx2 dx$$

$$displaystyle I=2pi ² - 2pi left[ tan dfracx2right]_0^pi $$

$$displaystyle I=2pi ² - 2pi left[ infty - 0 right] $$

$$I=-infty$$

Method 2

$$I=int_0^2pidfracxcos x1 +cos xdx.....(1)$$

$$I=int_0^2pidfrac(2pi-x)cos (2pi-x)1 +cos (2pi-x)dx.....(2)$$

Adding (1) and (2)

$$2I=int_0^2pidfrac2pi cos x1 +cos xdx$$

$$I=pi int_0^2pidfraccos x1 +cos xdx$$

$$I=piint_0^2pidfrac1+cos x-11 +cos xdx$$

$$I=pi int_0^2pi1-dfrac11 +cos xdx$$

$$I=pi int_0^2pi1dx-pi int_0^2pidfrac11 +cos xdx$$

$$I=pi[x]_0^2pi -pi int_0^2pidfrac12cos ²dfracx2dx$$

$$I=2pi ² - dfracpi2 int_0^2pi sec ²dfracx2 dx$$

$$displaystyle I=2pi ² - pi left[ tan dfracx2right]_0^2pi $$

$$displaystyle I=2pi ² - pi left[ 0-0right] $$

$$I=2pi ²$$

$$(P1)--> int_0^2a f(x)dx=2int_0^af(x)dx$$if $f(2a-x)=f(x)$

I have checked these methods several times, but can't find any mistake. Method 1 seems to be wrong as the given answer was $2pi ²$. What could be the mistake?

Actually, the first method is correct and the second method is wrong. To see what is happening here, notice that the integrand is almost proportional to $frac1(x-pi)^2$, precisely in the following sense:

$$ lim_x to pi (x-pi)^2 fracx cos x1+cos x
= -2pi. $$

So the integrand diverges to $-infty$ as $xtopi$.

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As far as improper integrability is concerned, having discontinuity points at which $|f(x)|$ diverges to $infty$ is not necessarily a threat for the convergence of the improper integral $int f(x) , dx$. In your case, however, we know that$int_pi-delta^pi+delta frac1(x-pi)^2 , dx = infty$ for any $delta > 0$, hence the integral in question does not converge by comparison test.

To see what is wrong in your second approach, note that $u = tan(x/2)$ has discontinuity at $x = pi$, hence cannot be used as a valid substitution over the interval $[0, 2pi]$. (Of course, it can be used as a substitution over the interval $[0,pi)$, which is why your first approach is valid.)

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