Find # int e^(-theta) cos8theta d theta #?
1 Answer
# int \ e^(-theta) \cos8theta \ d theta = e^(-theta)/65(8sin8theta -cos8theta) + C #
Explanation:
Let:
# I = int \ e^(-theta) \cos8theta \ d theta #
We can use integration by parts:
Let
# { (u,=cos8theta, => (du)/dx,=-8sin8theta), ((dv)/dx,=e^(-theta), => v,=-e^theta ) :}#
Then plugging into the IBP formula:
# int \ (u)((dv)/(d theta)) \ d theta = (u)(v) - int \ (v)((du)/(d theta)) \ d theta #
gives us
# int \ (cos8theta)(e^(-theta)) \ dx = (cos8theta)(-e^(-theta)) - int \ (-e^(-theta))(-8sin8theta) \ d theta #
# :. I = -e^(-theta)cos8theta - 8int \ e^(-theta) \ sin8theta \ d theta # .... [A]
At first it appears as if we have made no progress, as now the second integral is similar to
Let
# { (u,=sin8theta, => (du)/(d theta),=8cos8theta), ((dv)/(d theta),=e^(-theta), => v,=-e^(-theta) ) :}#
Then plugging into the IBP formula, gives us:
# int \ (sin8theta)(e^(-theta)) \ dx = (sin8theta)(-e^(-theta)) - int \ (-e^(-theta))(8cos8theta) \ d theta #
# :. int \ e^(-theta) \ sin8theta \ dx = -e^(-theta)sin8theta + 8I #
Inserting this result into [A] we get:
# I = -e^(-theta)cos8theta -8{-e^(-theta)sin8theta + 8I} + A #
# \ \ = -e^(-theta)cos8theta +8e^(-theta)sin8theta -64I + A #
# :. 65I = e^(-theta)(8sin8theta -cos8theta) + A #
# :. I = e^(-theta)/65(8sin8theta -cos8theta) + A/65 #
Or:
# I = e^(-theta)/65(8sin8theta -cos8theta) + C #