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Evaluate the following: ed∫e-3xcos3x dx - Mathematics

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Question

Evaluate the following:

`int "e"^(-3x) cos^3x  "d"x`

Sum
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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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Chapter 7: Integrals - Exercise [Page 166]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 42 | Page 166

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