Question

\(\int\limits^{\dfrac{\pi}{2}}_{\dfrac{\pi}{3}}\dfrac{1+sinx}{1+cosx}e^xdx\)

Answer 1

Ok bat ong doi lau roi

\(\int\dfrac{1+\sin x}{1+\cos x}e^xdx=\int\dfrac{e^xdx}{1+\cos x}+\int\dfrac{e^x\sin x}{1+\cos x}dx\)

\(I_1=\int\dfrac{e^xdx}{1+\cos x}\)

\(I_2=\int\dfrac{e^x\sin x}{1+\cos x}dx\)

\(\left\{{}\begin{matrix}u=\dfrac{\sin x}{1+\cos x}\\dv=e^xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{\cos x\left(1+\cos x\right)+\sin^2x}{\left(1+\cos x\right)^2}dx=\dfrac{dx}{1+\cos x}\\v=e^x\end{matrix}\right.\)

 \(\Rightarrow I_2=\dfrac{e^x.\sin x}{1+\cos x}-\int\dfrac{e^xdx}{1+\cos x}=\dfrac{e^x\sin x}{1+\cos x}-I_1\)

\(\Rightarrow I=\dfrac{e^x\sin x}{1+\cos x}\)

P/s: Done, ông thay cận vô nhé :)