\(I=\int_0^{\frac{\pi}{2}}\dfrac{e^x\sin x}{1+\sin 2x}dx\\ J=\int_0^{\frac{\pi}{2}}\dfrac{e^x\cos x}{1+\sin 2x}dx\)
\(\Rightarrow I-J=\int_0^{\frac{\pi}{2}}\dfrac{e^x(\sin x-\cos x)}{(\sin x+\cos x)^2}dx=\dfrac{e^x}{\sin x+\cos x}\Big|_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\dfrac{e^x}{\sin x+\cos x}dx\)
Suy ra
\(I-J=e^{\frac{\pi}{2}}-1-(I+J)\Rightarrow I=\dfrac{e^{\frac{\pi}{2}}-1}{2}\)
Tính \(\int_0^{\frac{\pi}{2}}\frac{sin2x}{3+4sinx-cos2x}dx\)
\(\int_0^{\dfrac{\pi}{2}}\)\(\dfrac{1+sin2x}{sinx+cosx}dx\)
\(\int_0^{\dfrac{\pi}{6}}\)\(\dfrac{1-sin2x+cos2x}{sinx-cos2x}dx\)
cho hàm số y=f(x) liên tục trên [0;π/2] thỏa \(\int_0^{\frac{\pi}{2}}f^2\left(x\right)dx=3\pi\) , \(\int_0^{\pi}\left(\sin x-x\right)f'\left(\frac{x}{2}\right)dx=6\pi\) ; \(f\left(\frac{\pi}{2}\right)=0\) Tính \(\int_0^{\frac{\pi}{2}}\left(f''\left(x\right)\right)^3dx\)
giúp em với ạ.
33/002
Tìm tích phân \(\int_0^{\frac{\pi}{2}}e^{-2cosx}dx\)
\(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln\left(\tan x\right)}{\sin2x}dx\)
Tính tích phân :
\(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{3}}\frac{\ln\left(4\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx\)
Cho f(x) liên tục trên R thỏa mãn \(\int_0^{\frac{1}{2}}f\left(\sqrt{1-2x^2}\right)dx\) = \(\frac{7}{6}\) và f (\(\frac{1}{\sqrt{2}}\)) =1. Tính I = \(\int_0^{\frac{\Pi}{4}}f'\left(cosx\right)sin^2xdx\)
A. \(\frac{1}{2}\) B.\(\frac{\sqrt{2}}{3}\) C. \(\frac{2\sqrt{2}}{3}\) D. 1
\(\int_0^{ln2}\frac{e^{2x}+3e^x}{e^{2x}+3e^x+2}dx\)
