Question

[Truongduy]

Biết rằng f(1/2)=a ;f(√3/2)=b và x + xf'(x)=2f(x) -4 .tính tích phân cận từ (π/6 đến π/3) của biểu thức[ (sinx^2*cosx + 2sin2x)]/[f(sinx)]^2 theo a và b

Answer 1

[Nguyễn Việt Lâm]

\(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{sin^2x.cosx+2sin2x}{\left(f\left(sinx\right)\right)^2}dx=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\left(sin^2x+4sinx\right).cosx}{\left(f\left(sinx\right)\right)^2}dx\)

Đặt \(sinx=t\Rightarrow cosx.dx=dt;\left\{{}\begin{matrix}x=\frac{\pi}{6}\Rightarrow t=\frac{1}{2}\\x=\frac{\pi}{3}\Rightarrow t=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow I=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\frac{\left(t^2+4t\right)}{f^2\left(t\right)}dt=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\frac{\left(x^2+4x\right)}{f^2\left(x\right)}dx\)

Lại có:

\(x+x.f'\left(x\right)=2f\left(x\right)-4\Leftrightarrow x+4=2f\left(x\right)-x.f'\left(x\right)\)

\(\Leftrightarrow x^2+4x=2x.f\left(x\right)-x^2.f'\left(x\right)\)

\(\Leftrightarrow\frac{x^2+4x}{f^2\left(x\right)}=\frac{2x.f\left(x\right)-x^2.f'\left(x\right)}{f^2\left(x\right)}=\left(\frac{x^2}{f\left(x\right)}\right)'\)

\(\Rightarrow I=\int\limits^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\left(\frac{x^2}{f\left(x\right)}\right)'dx=\frac{x^2}{f\left(x\right)}|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}=\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{f\left(\frac{\sqrt{3}}{2}\right)}-\frac{\left(\frac{1}{2}\right)^2}{f\left(\frac{1}{2}\right)}=\frac{3}{4b}-\frac{1}{4a}\)