Question

Evaluate

${\int }_0^{1}tan\left(\frac{2x-1}{1+x-x^2}\right)dx$

Answer 1

We have,

${\int }_0^1\tan \left(\frac{2x-1}{1+x-x^2}\right)dx$

Let

$1+x-x^2=t$

$\Rightarrow 0+1-2x=\frac{dt}{dx}$

$\Rightarrow (1-2x)dx=dt$

$\Rightarrow -(2x-1)dx=dt$

Change limit

If x=0 then,

$t=1+x-x^2$

$t=1+0-0^2$

$t=1$

If x=1 then,

$1+x-x^2=t$

$\Rightarrow 1+1-1^2=t$

$\Rightarrow t=1$

Now,

${\int }_1^1\tan \frac{1}{t}dt$

On integrating and we get,

$\Rightarrow {{\left[\log \sec t\right]}_1}^1$

$\Rightarrow [\log \sec 1-\log \sec 1]$

$\Rightarrow 0$

Hence, this is the answer.