Question

Find: $\int e^x\sin x.dx$

Answer 1

Let $I_1=\int \stackrel{\underset{}{e^x}}{II}.\stackrel{\underset{}{\sin x}}{I}.dx$
Using by parts
$\Rightarrow I_1=\sin x\int e^x.dx-\int (\frac{d\left(\sin x\right)}{dx}\int e^x.dx).dx$
$\Rightarrow I_1=e^x\sin x-\stackrel{\underset{}{\int \cos x.e^x.dx}}{I_2}$
Solving $I_2$
$I_2=\int \cos x.e^x.dx$
$\Rightarrow I_2=\cos x\int e^x.dx-\int (\frac{d\left(\cos x\right)}{dx}\int e^{x}dx).dx$
$\Rightarrow I_2=\cos x{e}^x-\int (-\sin x)e^x.dx$
$\Rightarrow I_2=e^x\cos x+\int \sin x{e}^x.dx$
$\Rightarrow I_2=e^c\cos x+I_1+C$
Now, putting value of $I_2$
$I-1=e^x\sin x-I_2$
$\Rightarrow I_1=e^x\sin x-(e^x\cos x+I_1)+C$
$\Rightarrow I_1=e^x\sin x-e^x\cos x-I_1+C$
$\Rightarrow 2I_1=e^x\sin x-e^x\cos x+C$
$\therefore I_1=\frac{e^x}{2}(\sin x-\cos x)+C$
Where $C$ is constant of integration.