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प्रश्न
Integrate the following with respect to x:
`"e"^("a"x) cos"b"x`
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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