Mathematics - Definite Integration Question with Solution | TestHub

$$I={\int }_0^{2\pi }(2\pi -x)\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

Using the property ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, we have:

$$I={\int }_0^{2\pi }(2\pi -(2\pi -x))\ln \left(\frac{3+\cos (2\pi -x)}{3-\cos (2\pi -x)}\right)dx$$

$$I={\int }_0^{2\pi }x\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

Adding the two expressions for $I$:

$$2I={\int }_0^{2\pi }(2\pi -x)\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx+{\int }_0^{2\pi }x\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

$$2I={\int }_0^{2\pi }(2\pi -x+x)\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

$$2I=2\pi {\int }_0^{2\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

Let $f(x)=\ln \left(\frac{3+\cos x}{3-\cos x}\right)$. We observe that $f(2\pi -x)=\ln \left(\frac{3+\cos (2\pi -x)}{3-\cos (2\pi -x)}\right)=\ln \left(\frac{3+\cos x}{3-\cos x}\right)=f(x)$.

Also, $f(x)$ is an odd function about $x=\pi$ in the interval $[0,2\pi ]$.

Consider the integral ${\int }_0^{2\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$.

Let $g(x)=\ln (3+\cos x)-\ln (3-\cos x)$.

Then $g(2\pi -x)=\ln (3+\cos (2\pi -x))-\ln (3-\cos (2\pi -x))=\ln (3+\cos x)-\ln (3-\cos x)=g(x)$.

Since $g(x)$ is an even function about $x=\pi$, we can write:

$${\int }_0^{2\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=2{\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

Now, let $x=\pi -t$. Then $dx=-dt$. When $x=0$, $t=\pi$. When $x=\pi$, $t=0$.

$${\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx={\int }_{\pi }^0\ln \left(\frac{3+\cos (\pi -t)}{3-\cos (\pi -t)}\right)(-dt)$$

$$={\int }_0^{\pi }\ln \left(\frac{3-\cos t}{3+\cos t}\right)dt$$

$$={\int }_0^{\pi }-\ln \left(\frac{3+\cos t}{3-\cos t}\right)dt$$

So, ${\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=-{\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$.

This implies $2{\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=0$.

Therefore, ${\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=0$.

Substituting this back into the expression for $2I$:

$$2I=2\pi {\int }_0^{2\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=2\pi \left(2{\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx\right)=2\pi (2\times 0)=0$$

Thus, $2I=0$, which means $I=0$.