Question

Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`

Answer 1

Let I = `int_0^(pi/4) (dx)/(1 + tanx)`

= `int_0^(pi/4) (dx)/(1 + sinx/cosx)`

= `int_0^(pi/4) (cos x dx)/(cosx + sinx)`

= `1/2 int_0^(pi/4) (2cosx)/(cosx + sinx) dx`

= `1/2 int_0^(pi/4) (cosx + sinx + cosx - sinx)/(cosx + sinx) dx`

= `1/2 [int_0^(pi/4) (cosx + sinx)/(cosx + sinx) dx + int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx]`

= `1/2 [int_0^(pi/4) 1dx + int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx]`

= `1/2 (I_1 + I_2)`

Where, I₁ = `int_0^(pi/4) 1dx`

= `[x]_0^(pi/4) = pi/4`

And I₂ = `int_0^(pi/4) (cosx - sinx)/(cosx + sinx) dx`

Let cosx + sinx = t

⇒ (–sinx + cosx)dx = dt

When x = 0, t = 1

And x = `pi/4`, t = `2/sqrt(2)`

∴ I₂ = `int_1^(2/sqrt(2)) (dt)/t`

= `[logt]_1^(2/sqrt(2))`

= `log  2/sqrt(2) - log 1`

= `log  2/sqrt(2) - 0`

= `log2^(3/2)`

= `3/2 log 2`

∴ I = `1/2(I_1 + I_2)`

or I = `1/2(pi/4 + 3/2 log 2)`