\(\int_0^{\dfrac{\pi}{2}}\)\(\dfrac{1+sin2x}{sinx+cosx}dx\)
\(\int_0^{\dfrac{\pi}{4}}\)\(\dfrac{cos2x-3sin^2x}{cos^2x}dx\)
Nguyên hàm từ 0 đến pi/6 của (1-sin2x+cos2x)/(sinx-cosx)dx
Tính \(\int_0^{\frac{\pi}{2}}\frac{sin2x}{3+4sinx-cos2x}dx\)
\(\int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}}\)\(\dfrac{cos2x}{sin^2x}dx\)
\(\int\limits^{\frac{Π}{2}}_{\frac{Π}{6}}\frac{1+SIN2x+cOS2x}{sINx+cosx}dx\)
a) \(\int_{\dfrac{\pi}{8}}^{\dfrac{2\pi}{8}}\)\(\dfrac{dx}{sin^2xcos^2x}\)
b) \(\int_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\)\(\dfrac{cos2xdx}{sin^2xcos^2x}\)
c) \(\int_0^{\dfrac{\pi}{3}}\)\(\dfrac{cos3x}{cosx}\)dx
\(\int_0^{\dfrac{\pi}{4}}\) \(\dfrac{3sin^2x-4cos^2x}{cos^2x}dx\)
Tính tích phân sau:
\(I=\int_0^{\pi}\dfrac{x.sinx}{sin^2x+3}dx\)