Question

\(\int_0^{\frac{\pi}{2}}\frac{e^xsinx}{1+sin2x}dx\)

Answer 1

\(I=\int_0^{\frac{\pi}{2}}\dfrac{e^x\sin x}{1+\sin 2x}dx\\ J=\int_0^{\frac{\pi}{2}}\dfrac{e^x\cos x}{1+\sin 2x}dx\)

\(\Rightarrow I-J=\int_0^{\frac{\pi}{2}}\dfrac{e^x(\sin x-\cos x)}{(\sin x+\cos x)^2}dx=\dfrac{e^x}{\sin x+\cos x}\Big|_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\dfrac{e^x}{\sin x+\cos x}dx\)

Suy ra

\(I-J=e^{\frac{\pi}{2}}-1-(I+J)\Rightarrow I=\dfrac{e^{\frac{\pi}{2}}-1}{2}\)