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"#
