If $f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in N$, then $f(t+\pi)=$

$$ \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{\frac{5}{2}} d x= $$

$$ \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x= $$

$$ \int_{1 / 5}^{1 / 2} \frac{\sqrt{x-x^2}}{x^3} d x= $$