Question

Evaluate: 

`int_0^(pi/4)(dx)/(cos^3 x sqrt(2 sin 2x))`

Answer 1

`int_0^(pi/4)(dx)/(cos^3 x sqrt(2 sin 2x))`

= `int_0^(pi//4)(dx)/(cos^3 x sqrt(2 xx 2 sin x cos x))`

= `int_0^(pi//4)(dx)/(2cos^3 x sin^(1/2)x cos^(1/2)x)`

∴ I = `int_0^(pi//4)dx/(2cos^(7/2) x sin^(1/2) x)`

Dividing by cos⁴ x in the numerator and denominator 

I = `int_0^(pi//4) ((1// cos^4 x  dx)/(2 cos^(7/2) x  sin^(1/2) x))/(cos^4 x)`

= `int_0^(pi//4)(sec^4 x  dx)/(2sqrt(tan x))`

= `int_0^(pi//4)(sec^2 x.  sec^2 x  dx)/(2sqrt(tan x))`

= `int_0^(pi//4)((1 + tan^2 x)sec^2 x)/(2sqrt(tan x))  dx`

Put tan x = t²

sec² x dx = 2t dt

When x = 0, t = `sqrt(tan 0) = 0`

`x = pi/4, t = sqrt(tan  pi/4 = 1)`

I = `int_0^1((1 + t^4) 2t)/(2t)  dt`

= `[t + (t^5)/(5)]_0^1`

= `(1 + 1/5) - 0`

= `((5 + 1)/5) - 0`

= `6/5`