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Evaluate the following : ∫e2x.cos3x.dx

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प्रश्न

Evaluate the following : `int e^(2x).cos 3x.dx`

योग
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उत्तर

Let I = `int e^(2x).cos 3x.dx`

I = `int cos 3x.e^(2x) dx`

= `cos 3x inte^(2x) .dx - int [d/dx (cos 3x) - e^(2x).dx]dx`

= `cos3x. (e^(2x))/(2) - int(-sin3x).(3) e^(2x)/2.dx`

= `(1)/(2).cos3xe^(2x) + 3/2 int sin 3x. e^(2x) dx`

= `(1)/(2)cos3xe^(2x) + 3/2[sin3x.int e^(2x)dx - int [(cos3x)3.int e^(2x)dx]dx`

= `(1)/(2)cos3x.e^(2x) + 3/2sin3x.(e^(2x))/2 - 3/2 .3int cos3x.e^(2x)/2dx`

= `(1)/(2)cos3x.e^(2x) + 3/4sin3x.e^(2x) - 9/4 intcos3x.e^(2x)dx`

= `(1)/(2)cos3x.e^(2x) + 3/4sin3x.e^(2x) - 9/4 "I"`

`"I" + 9/4"I" = (1/2 cos3x + 3/4 sin3x)e^(2x)`

`13/4"I" = (1/2 cos3x + 3/4 sin3x)e^(2x)`

I = `4/13 [1/2cos3x + 3/4sin3x]e^(2x)`

I = `1/13 [2cos3x + 3sin3x]e^(2x) + c`

∴ I = `e^(2x)/(13) (2 cos3x + 3 sin 3x) + c`.

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अध्याय 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.3 | Q 1.10 | पृष्ठ १३७

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