Evaluate: ??e^(-x) cos2xcos4xdx?
Asked by Manoj | 20th Mar, 2013, 05:09: AM
I = ?e^(-x) cos2xcos4x dx
Using the formula: 2cos x cos y = cos (x + y) + cos (x y)
I = 1/2?e^(-x) [cos6x + cos2x] dx
= 1/2?e^(-x) cos6x dx + 1/2?e^(-x) cos2x dx
= I1 + I2
Now, you can easily integrate the integrals on the RHS by integrating by parts.
Answered by | 22nd Mar, 2013, 01:49: PM
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