Question

Let $I={\int }_{3n\pi }^{(n+\frac{1}{n})3\pi }\frac{4xdx}{{[(a^2+b^2)+(a^2-b^2)\cos \frac{2nx}{3}]}^2}$ (where a,b>0)
prove that $I=\frac{9(2n^2+1)\pi }{n^2}\frac{a^2-b^2}{a^3{b}^3}$

Answer 1

Substitue $\frac{nx}{3}=t\Rightarrow \frac{n}{3}dx=dt$
$\therefore I={\int }_{n^2\pi }^{(n^2+1)\pi }\frac{\left(4.\frac{3t}{n}\right)\left(\frac{3}{n}dt\right)}{{[a^2(1+\cos 2t)+b^2(1-\cos 2t)]}^2}=\frac{9}{n^2}{\int }_{n^2\pi }^{(n^2+1)\pi }\frac{4tdt}{4{[a^2{\cos }^{2}t+b^2{\sin }^{2}t]}^2}$
$=\frac{9}{n^2}{\int }_{n^2\pi }^{(n^2+1)n}\frac{tdt}{{[a^2{\cos }^{2}t+b^2{\sin }^{2}t]}^2}$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I=\frac{9}{n^2}{\int }_{n^2\pi }^{(n^2+1)\pi }\frac{(2n^2+1)\pi -t}{{[a^2{\cos }^{2}t+b^2{\sin }^{2}t]}^2}$ ...(2)
Adding (1) and (2)
$\therefore 2I=\frac{9}{n^2}(2n^2+1)\pi .{\int }_{n^2\pi }^{(n^2+1)\pi }\frac{dt}{{[a^2{\cos }^{2}t+b^2{\sin }^{2}t]}^2}$
$\therefore I=\frac{9(2n^2+1)\pi }{n^2}\frac{a^2+b^2}{a^3{b}^3}$