integral calculation mistake I try to solve this:
$
\int_{0}^{\pi /2}\sin x \cos x\sqrt{1+\cos^{2}x }  dx
$
This is what I do:
$ \cos x = t; -\sin x dx = dt; -\sqrt{1-u^{2}} dt $
$ \frac{-\sqrt{1-t^{2}}*t*\sqrt{1+t^{2}}}{-\sqrt{1-t^{2}}} dt $
$-\int_{0}^{1} t * \sqrt{1+t^{2}} dt$
$1+t^2 = a; 2tdt = da; tdt = da/2$
$-\int_{0}^{2}\sqrt{a}da = - \frac{2\sqrt{2}}{3}$
But the answer is $\frac{2\sqrt{2}}{3} - \frac{1}{3}$
Where do I make the mistake

Your second substitution is: $a=1+t^2$. If $t \in [0,1]$, then $a \in [1,2]$ and not $a \in [0,2].$
