1
$\begingroup$
  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?

$\endgroup$
  • $\begingroup$ Please don't post images, especially low quality ones. Please use MathJax instead and type images out. $\endgroup$ – Em. Apr 8 '17 at 6:16
  • $\begingroup$ Got it. Wondered what everyone was using. $\endgroup$ – Ted Kumagai Apr 8 '17 at 6:56
2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Perfect! Thanks! Easier than I thought! $\endgroup$ – Ted Kumagai Apr 8 '17 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.