Question

The value of the integral `int_0^1 x cot^-1(1 - x^2 + x^4)dx` is ______.

Options

• `π/2 - 1/2 log_e 2`

• `π/4 - log_e 2`

• `π/2 - log_e 2`

• `π/4 - 1/2 log_e 2`

Answer 1

The value of the integral `int_0^1 x cot^-1(1 - x^2 + x^4)dx` is `underlinebb(π/4 - 1/2 log_e 2)`.

Explanation:

`int_0^1 x cot^-1(1 - x^2 + x^4)dx = int_0^1 xtan^-1(1/(1 + x^4 - x^2))`

= `int_0^1 x tan^-1 ((x^2 - (x^2 - 1))/(1 + x^2(x^2 - 1)))dx`

= `1/2 int_0^1 1tan^-1 t^2 dt - 1/2 int_-1^0 1tan^-1.k  dk`

Put x² = t `\implies` 2xdx = dt in the first integral
and x² – 1 = k `\implies` 2xdx = dk in the second integral.

= `1/2 int_0^1 1tan^-1 tdt - 1/2 int_0^1 1 tan^-1 kdk`

= `1/2(t tan^-1 t|_0^1 -int_0^1 t/(1 + t^2) dt) - 1/2(k tan^-1 k|_0^1 - int_-1^0 k/(1 + k^2) dk)`

= `1/2(π/4 - (1/2 ln (1 + t^2)|_0^1) - 1/2(-π/4 - (1/2 ln(1 + k^2)|_-1^0))`

= `(π/8 - 1/4 ln 2) - ((-π)/8 - 1/4 10 - ln 2)`

= `π/4 - 1/2 ln 2`