Hàm số \(F\left(x\right)\) xác định trên khoảng \(\left(0;\dfrac{\pi}{2}\right)\) thỏa mãn \(F\left(\dfrac{\pi}{6}\right)=\dfrac{\pi}{2},F\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4},F\left(\dfrac{\pi}{3}\right)=\pi\) và tồn tại \(a,b\) sao cho \(F'\left(x\right)=\dfrac{a\sin^2x\cos^2x+b\sqrt{3}}{\sin^2x\cos^2x}\). Hàm số $F(x)$ là
\(9x-2\pi\). \(x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)-\frac{\pi}{12}\). \(x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)\). \(x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)+\frac{\pi}{6}\). Hướng dẫn giải:\(F\left(x\right)=\int\dfrac{a\sin^2x\cos^2x+b\sqrt{3}}{\sin^2x\cos^2x}\text{dx}=\int\left(a+\dfrac{b\sqrt{3}}{\sin^2x\cos^2x}\right)\text{dx}\)
\(=\int\left[a+\left(\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}\right)b\sqrt{3}\right]\text{dx}\)
\(=ax+\left(-\cot x+\tan x\right)b\sqrt{3}+C\)
Để thỏa mãn các điều kiện: \(F\left(\dfrac{\pi}{6}\right)=\dfrac{\pi}{2},F\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4},F\left(\dfrac{\pi}{3}\right)=\pi\) ta có:
\(\left\{\begin{matrix}a.\dfrac{\pi}{6}+\left(-\cot\dfrac{\pi}{6}+\tan\dfrac{\pi}{6}\right)b\sqrt{3}+C=\dfrac{\pi}{2}\\a\dfrac{\pi}{4}+\left(-\cot\dfrac{\pi}{4}+\tan\dfrac{\pi}{4}\right)b\sqrt{3}+C=\dfrac{\pi}{4}\\a\dfrac{\pi}{3}+\left(-\cot\dfrac{\pi}{3}+\tan\dfrac{\pi}{3}\right)b\sqrt{3}+C=\pi\end{matrix}\right.\)
Giải hệ theo a, b, C ta được: \(a=1,b=\dfrac{\pi}{3},C=0\).
Vậy \(F\left(x\right)=x+\dfrac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)\).