# Evaluate $\int_0^1\int_0^{\cos^{-1}y}(\sin x)\sqrt{1+\sin^2x}\,dxdy.$

Evaluate $$\int_0^1\int_0^{\cos^{-1}y}(\sin x)\sqrt{1+\sin^2x}\,dxdy.$$

Can anyone hint me how to start solving this? Or solve the whole thing if you're generous enough. :D

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Since we know that the integrand is independent of $y$, we can perform integration w.r.t. $y$ first. Hence the integration becomes $\int_0^{\pi/2}\int_0^{\cos{x}}(\sin x)\sqrt{1+\sin^2x}\,dydx$ then the integration becomes a lot easier.
To OP: Drawing a picture can be helpful when trying to figure out if changing the order of integration would be helpful. After the order change, this becomes a very simple $u$ substitution. – apnorton Dec 11 '12 at 13:36
If we change the integration from $dxdy$ to $dydx$, that would mean that the bounds will change, right? How would I solve for the new boundaries? – Rose Dec 11 '12 at 13:42
@Rose: If you draw a picture of the region, the boundary curves are $x=0$, $y=0$, and $x=\cos^{-1}y$ (or equivalently, $y=\cos x$). If we go with $dxdy$, then we go from $x=0$ on the left to $x=\cos^{-1}y$ on the right, and then go from our lower bound $y=0$ to our upper bound, which is determined in this case by the point of intersection of $x=0$ and $x=\cos^{-1}y$, namely $y=\cos 0=1$. (cont'd) – Cameron Buie Dec 11 '12 at 15:44
Write out the original region of integration: $0 \leq x \leq \arccos y$, $0 \leq y \leq 1$. Now we want to write this a different way. If $0 \leq y \leq 1$, then $\arccos y$ ranges from $\arccos 0 \pi/2$ to $\arccos 1 =0$. So the largest $x$ can be is $\pi/2$. So the bounds on $x$ are $0 \leq x \leq \pi/2$. Now since $y$ ranges in $[0,1]$ and this happens to be the range of $\cos x$ on $[0,\pi/2]$, we have $0 \leq y \leq \cos x$. – Gyu Eun Lee Dec 11 '12 at 15:45