Answer:
[tex]\dfrac{1}{2}e^x(\sin(x)-\cos(x))+C[/tex]
Step-by-step explanation:
[tex]\displaystyle \int e^x \sin(x) \:dx[/tex]
Apply integration by parts:
[tex]\displaystyle \int u \dfrac{dv}{dx}\:dx=uv-\int v\: \dfrac{du}{dx}\:dx[/tex]
[tex]\textsf{Let }u=e^x \implies \dfrac{du}{dx}=e^x[/tex]
[tex]\textsf{Let }\dfrac{dv}{dx}=\sin(x) \implies v=- \cos(x)[/tex]
[tex]\begin{aligned}\implies \displaystyle \int e^x \sin(x) \:dx & = -e^x \cos(x)-\int - \cos(x)\: e^x\:dx\\& = -e^x \cos(x)+\int \cos(x)\: e^x\:dx\\\end{aligned}[/tex]
[tex]\displaystyle \textsf{Apply integration by parts to }\int \cos(x)\: e^x\:dx:[/tex]
[tex]\textsf{Let }u=e^x \implies \dfrac{du}{dx}=e^x[/tex]
[tex]\textsf{Let }\dfrac{dv}{dx}=\cos(x) \implies v=\sin(x)[/tex]
[tex]\implies \displaystyle \int \cos(x) e^x \:dx & =e^x \sin(x)-\int e^x \sin(x)\:dx[/tex]
Therefore:
[tex]\begin{aligned}\implies \displaystyle \int e^x \sin(x) \:dx & = -e^x \cos(x)+\left(e^x \sin(x)-\int e^x \sin(x)\:dx\right)\\\\\implies 2\int e^x \sin(x) \:dx & = -e^x \cos(x)+e^x \sin(x)\\ & = e^x(\sin(x)-\cos(x))\\\\\implies\int e^x \sin(x) \:dx & = \dfrac{1}{2}e^x(\sin(x)-\cos(x))+C \end{aligned}[/tex]
