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`        what will be the  integration of e to the power x sin square x dx?`
one year ago

```							Let L = ∫ e^x*sin(2x) dx and use integration by parts. Let: u = sin(2x) du = 2cos(2x) dx dv = e^x dx v = e^x Then: L = uv - ∫ v du ==> L = e^x*sin(2x) - 2 ∫ e^x*cos(2x) dx Let: u = cos(2x) du = -2sin(2x) dv = e^x dx v = e^x By another round of integration by parts: L = e^x*sin(2x) - 2(uv - ∫ v du) ==> L = e^x*sin(2x) - 2[e^x*cos(2x) + 2 ∫ e^x*sin(2x) dx] ==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4 ∫ e^x*sin(2x) dx ==> L = e^x*sin(2x) - 2e^x*cos(2x) - 4L (since L = ∫ e^x*sin(2x) dx) ==> 5L = e^x*sin(2x) - 2e^x*cos(2x) ==> 5L = e^x * [sin(2x) - 2cos(2x)] ==> L = e^x * [sin(2x) - 2cos(2x)]/5 ==> ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 + C Therefore, ∫ e^x*sin(2x) dx = e^x * [sin(2x) - 2cos(2x)]/5 +
```
one year ago
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