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

integrationmultivariable-calculus

  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?

Best Answer

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?

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