Complex Integration Homework - Part (a) and (b) Help

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Homework Statement



attachment.php?attachmentid=68696&stc=1&d=1397618882.jpg


Homework Equations





The Attempt at a Solution



I did part (a) which is pretty easy. Use the Cauchy integral formula. Part (b) says "hence", which leads me to believe that part (a) can be used for part (b), but I cannot see anything remotely related between the two. I don't want to do weird manipulations and substitutions if part (a) can help.

If someone could help me out with this I would really appreciate it.
 

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stripes said:

Homework Statement



attachment.php?attachmentid=68696&stc=1&d=1397618882.jpg


Homework Equations





The Attempt at a Solution



I did part (a) which is pretty easy. Use the Cauchy integral formula. Part (b) says "hence", which leads me to believe that part (a) can be used for part (b), but I cannot see anything remotely related between the two. I don't want to do weird manipulations and substitutions if part (a) can help.

If someone could help me out with this I would really appreciate it.

Just substitute ##e^{i \theta}## for ##z## in the integral and turn it into an integral ##d\theta##. Then use deMoivre.
 
Thanks Dick! Easy as pie now
 
Here is an elementary approach for part b) if you like:

Notice that the integral you seek is
$$\begin{aligned}
\Re\left(\int_0^{\pi} e^{k\cos t}e^{ik\sin t}\,dt\right) &= \Re\left(\int_0^{\pi}e^{ke^{it}}\,dt\right)\\
&=\Re\left(\int_0^{\pi} \left(1+ke^{it}+k^2e^{2it}+\cdots \right)\,dt\right)\\
&=\boxed{\pi}\\
\end{aligned}$$
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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