# Evaluate `int e^(8theta) sin(9theta)d theta` .

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### 1 Answer

You should use integration by parts such that:

`int udv = uv - int vdu`

`u = e^(8theta) => du = 8e^(8theta)d theta`

`dv =sin 9 theta => v = -(cos 9theta)/9`

`int e^(8theta)*sin 9 theta dtheta =-e^(8theta)*(cos 9theta)/9+ (8/9) int e^(8theta)*cos 9 theta d theta`

You should solve `int e^(8theta)*cos 9 theta d theta ` using parts such that:

`u = e^(8theta) => du = 8e^(8theta)d theta`

`dv = cos 9 theta => v = (sin 9 theta)/9`

`int e^(8theta)*cos 9 theta d theta = e^(8theta)*(sin 9 theta)/9 - (8/9) int e^(8theta)*sin 9 theta d theta`

You should come up with the following notation `int e^(8theta)*sin 9 theta dtheta = I` such that:

`I = -e^(8theta)*(cos 9theta)/9+ (8/9)*(e^(8theta)*(sin 9 theta)/9 - (8/9)I)`

You should open the brackets such that:

`I = -e^(8theta)*(cos 9theta)/9+ (8e^(8theta)*(sin 9 theta))/81 - 64/81 I`

You need to move the terms that contain I to the left side such that:

`I + 64/81I = -e^(8theta)*(cos 9theta)/9 + (8e^(8theta)*(sin 9 theta))/81`

`145/81 I = -e^(8theta)*(cos 9theta)/9 + (8e^(8theta)*(sin 9 theta))/81 `

`I = (-9e^(8theta)*(cos 9theta) + 8e^(8theta)*(sin 9 theta))/145`

**Hence, evaluating the given integral using parts yields `int e^(8theta)*sin 9 theta d theta =(-9e^(8theta)*(cos 9theta) + 8e^(8theta)*(sin 9 theta))/145.` **

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