Can anyone help solve the information on this page?


Guest May 27, 2017

I do not know what this question is about but


\(cos2\theta=cos^2\theta-sin^2\theta \\ cos2\theta=cos^2\theta-(1-cos^2\theta) \\ \frac{cos2\theta+1}{2}=cos^2\theta\\\)




\(\displaystyle\int_0^{\pi/4} cos^2\theta\;d\theta \\ =\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)+1}{2}\;d\theta \\ =\frac{1}{2}\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)}{1}+1\;d\theta \\ =\frac{1}{2}\left[ \frac{sin(2\theta)}{2}+\theta\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ sin(2\theta)+2\theta\;\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ (sin(\pi)+\pi)-(0)\;\right ]\\ =\frac{\pi}{4}\)


I don't know if that helps or not.

Melody  May 27, 2017

As follows:


I.  Melody has shown that the result is pi/4.  This is 0.79 to two dp.


II.  Use the fact that \(e^{i\pi/2}=i\)   so \(i^i=(e^{i\pi/2})^i\rightarrow e^{i\times i\pi/2}\rightarrow e^{-\pi/2}\)


Hence \(\chi=5+ie^{-\pi/2}\)


\(Real(\chi)=5.00\) to two dp   and  \(Imag(\chi)=0.21\)to two dp


III.  Separate variables and use the fact that \(\frac{1}{\psi(1-\psi)}=\frac{1}{\psi}+\frac{1}{1-\psi}\)the integrals are then straightforward.  Remember to include a constant, k, say, because the integrals are indefinite.  You then have two unknowns, \(\beta\) and k.  Use the two given conditions to find them.  Then you can determine \(\psi(4)\).  (You should find \(\psi(4) = 0.99\) to two dp).



(I would have supplied more detail, but the image uploading process isn't working at present!)

Alan  May 27, 2017

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