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Integrate the following functions w.r.t. x : e2x.sin3x

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प्रश्न

Integrate the following functions w.r.t. x : `e^(2x).sin3x`

बेरीज
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उत्तर

Let I = `int e^(2x).sin3x`

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

I = `sin3x . int e^(2x) dx - int[d/dx (sin3x) int e^(2x)dx]dx`

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

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

I = `1/2sin3x.e^(2x) - 3/2 intcos3x inte^(2x)dx - int [d/dx cos3x . int e^(2x)dx]dx`

I = `1/2 sin3x . e^(2x) - 3/2 cos3x . e^(2x)/2 + 3/2 int -sin3x . x3 . e^(2x)/2 dx` 

I = `1/2 sin3x . e^(2x) - 3/4 cos3x . e^(2x) - 9/4 [int sin3x . e^(2x) dx]`

I = `1/2 sin3x . e^(2x) - 3/4 . cos3x . e^(2x) - 9/4 "I" + "c"_1`

`"I" + 9/4"I" = 1/2 sin3x . e^(2x) - 3/4 cos3x . e^(2x) + "c"_1`

`13/4 "I" = 1/2 e^(2x) [sin3x - 3/2 cos3x] + "c"_1`

I = `4/13 xx 1/2 e^(2x) [sin3x . 3/2 cos3x] + 4/13 "c"_1    ...[at  4/13 "c"_1 = "c"]`

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

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पाठ 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३८]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.3 | Q 2.01 | पृष्ठ १३८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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