Evaluate the$\int_0^1\int_{\sin^{-1} y}^{\pi/2} \cos x\sqrt{1+\cos^2}\,dxdy$ and $\int_0^2\int_0^1\int_y^1 \sinh(z^2)\,dzdydx$ 
  
*
  
*Evaluate the following integrals:
  \begin{gather}
\int_0^1\int_{\sin^{-1} y}^{\pi/2} \cos x\sqrt{1+\cos^2 x}\,dxdy;\\
\int_0^2\int_0^1\int_y^1 \sinh(z^2)\,dzdydx
\end{gather}
  



*

*I tried doing integration by parts, got nowhere as nothing seems to reduce. I tried converting $\cos(x)^2$ into $1-\sin(x)^2$ and u-sub, but ended up with $\sqrt{2-u^2}$. Again stuck.

*Know the hyperbolic equivalents $\sinh x = \frac{e^x-e^{-x}}{2}$, and evaluate the integral with $x=z^2$.
Help with number 1? Also maybe links to how to deal with these types of integrals?
 A: Help with number 1?
Hint. You are on the right track. To evaluate 
$$
\int \sqrt{2-u^2}du
$$ one may use the change of variable $u=\sqrt{2}\cdot \sin t$ obtaining
$$
\begin{align}
\int \sqrt{2-u^2}du&=2\int \cos^2 t \:dt
\\&=2\int \left(\frac12+\frac{\cos (2t)}2 \right)dt
\\&=\int \left(1+\cos (2t) \right)dt
\\&=t+\frac{\sin (2t)}2
\\&=t+\sin t \cdot \cos t
\\&=\arcsin\left(\frac{u}{\sqrt{2}}\right)+\frac{1}{2}u \sqrt{2-u^2}.
\end{align}
$$ Thus
$$
\begin{align}
\int_{\arcsin y}^{\pi/2}\cos x \cdot \sqrt{1+\cos^2 x}\:dx&=\int_{\arcsin y}^{\pi/2}\cos x \cdot \sqrt{2-\sin^2 x}\:dx
\\\\&=\int_{y}^{1}\sqrt{2-u^2 }\:du
\\\\&=\frac12+\frac{\pi}4-\frac12\cdot y \sqrt{2-y^2}-\arcsin\left(\frac{y}{\sqrt{2}}\right)
\end{align}
$$ Can you finish it?
A: The trick for integral number 2 is to switch the order of integration as integrating with respect to y is much easier than with respect to z.
$$\int_{0}^{1} \int_{y}^{1} sinh(z^2) dz dy = \int_{0}^{1} \int_{0}^{z} sinh(z^2) dy dz$$
Now evaluating the inner integral is no problem and the result is
$$\int_{0}^{1} z sinh(z^2) dz$$
Now substitute $u = z^2$ to finish off the problem.
