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

After your first substitution you should have $-\int_{1}^{0}t\sqrt{1+t^{2}}dt$= $\int_{0}^{1}t\sqrt{1+t^{2}}dt$.
The limits for your second substitution should be $a=1$ at ($t=0$) and $a=2$ (at $t=1$) so we have $\frac{1}{2}\int_{1}^{2} \sqrt{a}da=\frac{1}{3}a^{\frac{3}{2}}|^{a=2}_{a=1}=\frac{2\sqrt{2}}{3}-\frac{1}{3}$ as required.
