Question

Solve


`\int e^x sin(x)dx`

Answer 1

`(1)/(2)e^x(\sin(x)-\cos(x))+C`

Explanation:

`\displaystyle \int e^x \sin(x) \:dx`

Apply integration by parts:

`\displaystyle \int u (dv)/(dx)\:dx=uv-\int v\: (du)/(dx)\:dx`

`\textsf{Let }u=e^x \implies (du)/(dx)=e^x`

`\textsf{Let }(dv)/(dx)=\sin(x) \implies v=- \cos(x)`

`\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}`

`\displaystyle \textsf{Apply integration by parts to }\int \cos(x)\: e^x\:dx:`

`\textsf{Let }u=e^x \implies (du)/(dx)=e^x`

`\textsf{Let }(dv)/(dx)=\cos(x) \implies v=\sin(x)`

`\implies \displaystyle \int \cos(x) e^x \:dx & =e^x \sin(x)-\int e^x \sin(x)\:dx`

Therefore:

`\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 & = (1)/(2)e^x(\sin(x)-\cos(x))+C \end{aligned}`