+118608 I do not know what this question is about but
\(cos2\theta=cos^2\theta-sin^2\theta \\ cos2\theta=cos^2\theta-(1-cos^2\theta) \\ \frac{cos2\theta+1}{2}=cos^2\theta\\\)
so
\(\displaystyle\int_0^{\pi/4} cos^2\theta\;d\theta \\ =\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)+1}{2}\;d\theta \\ =\frac{1}{2}\displaystyle\int_0^{\pi/4} \frac{cos(2\theta)}{1}+1\;d\theta \\ =\frac{1}{2}\left[ \frac{sin(2\theta)}{2}+\theta\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ sin(2\theta)+2\theta\;\right ]_0^{\pi/2}\\ =\frac{1}{4}\left[ (sin(\pi)+\pi)-(0)\;\right ]\\ =\frac{\pi}{4}\)
I don't know if that helps or not.
+33614 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!)
