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Question
Evaluate the following:
`int "e"^(-3x) cos^3x "d"x`
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Solution
Let I = `int "e"^(-3x) cos^3x "d"x`
= `int "e"^(-3x) ((cos 3x + 3 cosx)/4) "d"x`
= `1/4 int ("e"^(-3x) cos 3x + "e"^(-3x) cos x) "d"x`
= `1/4 ("I"_1 + "I"_2)`
I1 = `int "e"^(-3x) cos 3x "d"x`
= `"e"^(-3x) int cos 3x "d"x - int (("e"^(-3x)) int cos 3x "d"x) "d"x`
= `"e"^(-3x) sin (3x)/3- int - 3"e"^(-3x) sin (3x)/3 "d"x`
= `"e"^(-3x) sin (3x)/3 + "e"^(-3x) sin 3x "d"x`
= `"e"^(-3x) sin (3x)/3 + "e"^(-3x) cos (3x)/3 - int (("e"^(-3x))"'" int sin 3 x "d"x)"d"x`
= `"e"^(-3x) sin (3x)/3 - "e"^(-3x) cos (3x)/3 - int "e"^(-3x) cos 3x "d"x`
= `"e"^(-3x) sin (3x)/3 - "e"^(-3x) cos (3x)/3 - "I"_1`
⇒ 2I = ("e"^(-3x))/3 (sin 3x - cos 3x)`
⇒ I1 = `("e"^(-3x))/6 (sin 3x - cos 3x) + "C"_1`
Similarly I2 = `int "e"^(-3x) cos x"d"x`
= `("e"^(-3x))/10 (sin 3x - 3 cos 3x) + "C"_2`
⇒ I = `1/4 [("e"^(-3x))/6 (sin 3x - cos 3x) + "e"^(-3x)/10 (sin 3x - 3 cos 3x)] + "C"`
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