# Evaluate the$\int_0^1\int_{\sin^{-1} y}^{\pi/2} \cos x\sqrt{1+\cos^2}\,dxdy$ and $\int_0^2\int_0^1\int_y^1 \sinh(z^2)\,dzdydx$

1. 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}
1. 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.

2. 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?

• Please don't post images, especially low quality ones. Please use MathJax instead and type images out. – Em. Apr 8 '17 at 6:16
• Got it. Wondered what everyone was using. – Ted Kumagai Apr 8 '17 at 6:56

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?