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Integrate the following with respect to x: eabeaxcosbx - Mathematics

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प्रश्न

Integrate the following with respect to x:

`"e"^("a"x) cos"b"x`

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उत्तर

`int "e"^("a"x) cos"b"x`

Let I = `int "e"^("a"x) cos "b"x  "d"x`

u = cos bx

⇒ dc = `"e"^("a"x)  "d"x`

du = – b sin b dx

⇒  = `"e"^("a"x)/"a"`

Applying Integration by parts,

I = `"e"^("a"x)/"a"  cos  "b"x - int "e"^("a"x)/"a" (- "b" sin "b"x)  "d"x`

I = `"e"^("a"x)/"a" cos "b"x + "b"/"a" int"e"^("a"x) sin "b"x  "d"x`

u = sin x bx

⇒ dv = `"e"^("a"x)  "d"x`

du = b cos bx dx

⇒  = `"e"^("a"x)/"a"`

Again Applying Integration by parts,

I = `"e"^("a"x)/"a" cos "b" + "b"/"a" ["e"^("a"x)/"a" sin "b"x - "b"/"a" int "e"^("a"x) cos bx  "d"x]`

I = `("e"^("a"x) cos "b"x)/"a" + "b"/"a" [("e"^("a"x) sin "b"x)/"a" - "b"/"a" "I"]`

I  = `("e"^("a"x) cos "b"x)/"a" + ("be"^("a"x) sin "b"x)/"a" - "b"^2/"a"^2 "I"`

`"I" + "b"^2/""^2 "I" = ("ae"^("a"x) cos"b"x + "be"^("a"x) sin"b"x)/"a"^2 + "c"`

`"I"(("a"^2 + "b"^2)/"a"^2) =  ("e"^("a"x) ["a"cos"b"x + "b" sin"b"x])/"a"^2 + "c"`

∴ `int "e"^("a"x) cos "b"x  "d"x =  "e"^("a"x)/("a"^2 + "b"^2) ["a" cos "b"x + "b" sin "b"x] + "c"``

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Integral Calculus - Exercise 11.8 [पृष्ठ २१२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 11 Integral Calculus
Exercise 11.8 | Q 1. (i) | पृष्ठ २१२

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