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Spoiler
Spoiler
\begin{array}{rl}[br]I+J & \displaystyle = \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \\ \br \\[br]& \displaystyle = \int_0^a 1 \, dx \\ \br \\[br]& \displaystyle = \left[ x \right]_0^a = a \end{array}
\begin{array}{rl}[br]I-J & \displaystyle = \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \\ \br \\[br]& \displaystyle = \left[ \ln (\sin x + \cos x) \right]_0^a \\ \br \\[br]& \displaystyle = \ln (\sin a + \cos a)\end{array}
(i)
(ii)
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\displaystyle \int_{\pi/6}^{\pi/4} \frac{1}{1-\cos{\2\theta}} \,\mathrm{d}\theta =
\displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}x\,\mathrm{d}x = \alphaWl^{-\alpha}\int_0^l x^\alpha \,\mathrm{d}x
Q \~ B(L,z)
\displaystyle Q' \~ N(Lz,Lz(1-z))
\displaystyle q = P(Q>l) \approx 1 - \Phi\left(\frac{(l-\frac{1}{2})-Lz}{\sqrtLz(1-z)}\right) = \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz(1-z)}}\right) \approx \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz}}\right)
\displaystyle[br]y^2 = \left\{ \begin{array}{ll}[br]1 - (x + \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\[br] & \\[br]1 + (x + \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\[br]\end{array}[br]
\displaystyle[br]y^2 = \left\{ \begin{array}{ll}[br]1 - (x - \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\[br] & \\[br]1 + (x - \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\[br]\end{array}[br]