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Question
Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "d"x`
Solution
Let I = `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "d"x`
= `int_0^(pi/4) ((sin^3x)/(cos^3x))/(2cos^2x) "d"x`
= `1/2int_0^(pi/4) (sin^2x)/(cos^5x) "d"x`
= `1/2 int_0^(pi/4) ((1 - cos^2x)sinx)/(cos^5x) "dx`
Put cos x = t
∴ − sin x dx = dt
∴ sin x dx = − dt
When x = 0, t =1 and when x = `pi/4`, t = `1/sqrt(2)`
∴ I = `1/2 int_1^(1/sqrt2) (-(1 - "t"^2))/"t"^5 "dt"`
= `1/2 int_1^(1/sqrt(2)) ("t"^2 - 1)/"t"^5 "dt"`
= `1/2 int_1^(1/sqrt(2)) ("t"^(-3) - "t"^(-5)) "dt"`
= `1/2 int_1^(1/sqrt(2)) "t"^(-3) "dt" - 1/2 int_1^(1/sqrt(2)) "t"^(-5) "dt"`
= `1/2[("t"^(-2))/(-2)]_1^(1/sqrt(2)) - 1/2[("t"^-4)/(-4)]_1^(1/sqrt(2))`
= `-1/4[1/("t"^2)]_1^(1/sqrt(2)) + 1/8[1/("t"^4)]_1^(1/sqrt(2))`
= `-1/4(1/(1/2)- 1) + 1/8(1/(1/4) - 1)`
= `-1/4 + 3/8`
∴ I = `1/8`
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