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Question
Solve the following:
`("d"y)/("d"x) + y tan x = cos^3x`
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Solution
It is of the form `("d"y)/("d"x) + "P"y` = Q
Here P = tan x
Q = cos3x
`int "Pd"x = int tan x "d"x`
= `int sinx/cosx "d"x`
= `- int (- sinx)/cosx "d"x`
= – log cos x
= log sec x
I.F = `"e"^(int Pdx)`
= `"e"^(log sec x)`
= sec x
The required solution is
y(I.F) = `int "Q" ("I.F") "d"x + "c"`
y(sec x) = `int cos^3x (sec x) "d"x + "c"`
y(sec x) = `int cos^3x 1/cosx "d"x + "c"`
y(sec x) = `int cos^2x "d"x + "c"`
y(sec x) = `int ((1 + cos 2x)/2) "d"x + "c"`
y(sec x) = `1/2 int (1 + cos2x) "d"x + "c"`
y(sec x) = `1/2 [x + (sin2x)/2] + "c"`
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