Given inte^(8theta)sin(9theta)d theta
We will Apply Integration by Parts (I.B.P) twice and solve the integral algebraically at the end:
int u_1 dv_1=uv-int v du
Also: We'll let inte^(8theta)sin(9theta)d theta=I
Let color(red)(u=e^(8theta);dv=sin(9theta)
Thus: color(red)(du_1=8e^(8theta)d theta;v_1=-1/9cos(9theta)
I=color(red)(e^(8theta)*-1/9cos(9theta)-int-1/9cos(9theta)*8e^(8theta)d theta
Simplify
I=color(red)(-1/9e^(8theta)cos(9theta)+8/9intcos(9theta)*e^(8theta)d theta
Apply I.B.P again
color(blue)(u_2=e^(8theta);dv_2=cos(9theta)
color(blue)(du_2=8e^(8theta)d theta;v_2=1/9sin(9theta)
I=color(red)(-1/9e^(8theta)cos(9theta)+8/9)color(blue)({e^(8theta)*1/9sin(9theta)-int1/9sin(9theta)*8e^(8theta)d theta}
Simplify
I=color(red)(-1/9e^(8theta)cos(9theta)+8/9)color(blue)({1/9e^(8theta)sin(9theta)-8/9intsin(9theta)*e^(8theta)d theta}
I=color(red)(-1/9e^(8theta)cos(9theta)+)color(blue)(8/81e^(8theta)sin(9theta)-64/81inte^(8theta)sin(9theta) d theta
**Notice how the integral ** inte^(8theta)sin(9theta)d theta appears both on the left and right sides and so we can simplify this algebraic expression simply by solving for inte^(8theta)sin(9theta)d theta
Add 64/81inte^(8theta)sin(9theta)d theta to both sides
145/81inte^(8theta)sin(9theta)d theta=-1/9e^(8theta)cos(9theta)+8/81e^(8theta)sin(9theta)
Divide 145/81 to both sides
inte^(8theta)sin(9theta)d theta=81/145{-1/9e^(8theta)cos(9theta)+8/81e^(8theta)sin(9theta)}
Simplify
inte^(8theta)sin(9theta)d theta=-81/1305e^(8theta)cos(9theta)+648/11745e^(8theta)sin(9theta)
inte^(8theta)sin(9theta)d theta=-9/145e^(8theta)cos(9theta)+8/145e^(8theta)sin(9theta)+"C"
Alternatively, we can factor out -1/145e^(8theta) to get:
inte^(8theta)sin(9theta) d theta=-1/145e^(8theta){9cos(9theta)-8sin(9theta)}+"C"
