Substitution in easy definite integral

Question

Let me ask you the following very easy question, but I have some problems. I want to show that $$\int \limits_{0}^{2\pi}\dfrac{d\theta}{a+\cos \theta}=2\int \limits_{0}^{\pi}\dfrac{d\theta}{a+\cos \theta}.$$

It's enough to show that $\int \limits_{\pi}^{2\pi}\dfrac{d\theta}{a+\cos \theta}=\int \limits_{0}^{\pi}\dfrac{d\theta}{a+\cos \theta}$.

Consider the integral on the LHS and make substitution $\theta=x+\pi$ and we get the following $\int \limits_{0}^{\pi}\dfrac{d(x+\pi)}{a+\cos (x+\pi)}=\int \limits_{0}^{\pi}\dfrac{dx}{a-\cos x}$ which is not I want.

Maybe i am doing smth wrong.

Can anyone point out what's wrong with my reasoning?

Answer 1

Because $$\int \limits_{0}^{\frac{\pi}{2}}\dfrac{d\theta}{a+\cos \theta}=\int \limits_{2\pi}^{\frac{3\pi}{2}}\dfrac{d\left(2\pi-\theta\right)}{a+\cos(2\pi-\theta)}=\int \limits_{\frac{3\pi}{2}}^{2\pi}\dfrac{d\theta}{a+\cos \theta}$$ and $$\int \limits_{\frac{\pi}{2}}^{\pi}\dfrac{d\theta}{a+\cos \theta}=\int \limits_{\pi}^{\frac{3\pi}{2}}\dfrac{d\theta}{a+\cos \theta}.$$

Answer 2

Hint: Substitute $$\cos(x)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2dt}{1+t^2}dt$$ the so-called Weierstrass Substitution.