Đáp án: $B$
Giải thích các bước giải:
Ta có:
$F(x)=\displaystyle\int\dfrac{(a-b)\sin^2x+b}{\sin^2x\cos^2x}dx$
$\to F(x)=\displaystyle\int\dfrac{(a-b)\sin^2x+b(\sin^2x+\cos^2x)}{\sin^2x\cos^2x}dx$
$\to F(x)=\displaystyle\int\dfrac{a\sin^2x+b\cos^2x}{\sin^2x\cos^2x}dx$
$\to F(x)=\displaystyle\int\dfrac{a}{\cos^2x}+\dfrac{b}{\sin^2x}dx$
$\to F(x)=a\tan x-b\cot x+C$
Mà $F(\dfrac{\pi}{4})=\dfrac12, F(\dfrac{\pi}{6})=0, F(\dfrac{\pi}{3})=1$
$\to \begin{cases}a\tan(\dfrac{\pi}{4})-b\cot(\dfrac{\pi}{4})+C=\dfrac12\\a\tan(\dfrac{\pi}{6})-b\cot(\dfrac{\pi}{6})+C=0\\a\tan(\dfrac{\pi}{3})-b\cot(\dfrac{\pi}{3})+C=1\end{cases}$
$\to \begin{cases}a-b+C=\dfrac12\\\dfrac{\sqrt{3}a}{3}-\sqrt{3}b+C=0\\\sqrt{3}a-\dfrac{\sqrt{3}b}{3}+C=1\end{cases}$
$\to a=b=\dfrac{\sqrt{3}}{4}, C=\dfrac12$
$\to F(x)=\dfrac{\sqrt{3}}{4}\tan x-\dfrac{\sqrt{3}}{4}\cot x+\dfrac12$
$\to F(x)=\dfrac{\sqrt{3}}{4}(\tan x-\cot x)+\dfrac12$
$\to B$
