$f(x)=\int\limits_{\cos x}^{\sin x}\left(1-t+2 t^3\right) d t$ has in $[0,2 \pi]$

a maximum at $\frac{3 \pi}{4}$ and a minimum at $\frac{7 \pi}{4}$

We have,

$f(x)=\int\limits_{\cos x}^{\sin x}\left(1-t+2 t^3\right) d t$

$\Rightarrow f'(x) =\cos x\left(1-\sin x+2 \sin ^3 x\right) +\sin x\left(1-\cos x+2 \cos ^3 x\right)$          [Using Leibnitz's Rule]

$\Rightarrow f'(x)=\cos x+\sin x-2 \sin x \cos x$

$\Rightarrow f'(x)=\cos x+\sin x \cos x\left(\sin ^2 x+\cos ^2 x\right)$

For maximum or minimum, we have

$f'(x)=0 \Rightarrow \tan x=-1 \Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$

Now,

$f''(x)=-\sin x+\cos x$

∴  $f''\left(\frac{3 \pi}{4}\right)<0$ and $f''\left(\frac{7 \pi}{4}\right)>0$

Hence, f(x) attains a local maximum at $x=\frac{3 \pi}{4}$ and a minimum at $x=\frac{7 \pi}{4}$.