How do I improve my approach to solving integrals to get this and similar ones in the future correct? $$\int \sqrt{ 8 (\cos t \sin t)^2 } dt = \sqrt{2} \int 2\sin t\cos t dt = \sqrt{2} (\sin t)^2 + C$$
Which seems correct to me, but if I take the definite integral from $0$ to $\pi$, then:
$$\sqrt{2} \left( (\sin \pi)^2 - (\sin 0)^2 \right) = \sqrt{2} ( 0 - 0 ) = 0$$
But my solution manual says $2 \sqrt{2}$, and wolframalpha also gives that answer.
 A: $\LARGE \int \sqrt{8(\cos t \sin t)^2} dt$
$\LARGE\sqrt 2\int \sqrt{(2 \sin t \cos t)^2} dt$
$\LARGE \sqrt 2 \int \sqrt{(\sin 2t)^2} dt$
$\LARGE \sqrt 2 \int |\sin 2t| dt$
Here if you observe, $\Large \sin 2t$ is faster or has more frequency than $\Large \sin t$, so by the time $\Large \sin t$ has traversed half a graph, $\Large \sin 2t$ has traversed full graph which implies $\Large \sin 2t $ has a period of $\Large \pi$ (half of that of $\sin t$), so if you integrate $\Large \sin 2t$ over a period of $\Large0$ to $\Large \pi$, you will get answer as zero as the area between the X-axis and the $\Large \sin 2t$ graph below the the X-axis cancels out the area above the X-axis lying between X-axis and $\Large \sin 2t$ graph.  So, you integrate the function over half period and then multiply the value by $\color{red}{two}$.
$\LARGE \sqrt 2 . \color{red}{2} \int^{\pi/2}_0 \sin 2t dt$
$\LARGE \sqrt 2 (- \cos \pi - \cos 0)$
$\LARGE2 . \sqrt 2$
I hope this helps, still cross check it with someone.
A: Some graphs to contemplate:${}{}{}$

