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Question
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Solution
Let I = `int_0^(pi/4) sec^4x "d"x`
= `int_0^(pi/4) sec^2x*sec^2 x "d"x`
= `int_0^(pi/4) (1 + tan^2x)sec^2 x "d"x`
Put tan x = t
∴ sec2x dx = dt
When x = 0, t = 0 and when x = `pi/4`, t = 1
∴ I = `int_0^1(1 + "t"^2)"dt"`
= `int_0^1 "dt" + int_0^1 "t"^2 "dt"`
= `["t"]_0^1 + ["t"^3/3]_0^1`
= `(1 - 0) + 1/3(1^3 - 0)`
= `4/3`
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