Question

Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x)  "d"x`

Answer 1

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`