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\displaystyle \psi=\int^{\beta}_{\alpha}{\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,dx
\displaystyle \omega=\int^{\beta}_{\alpha}{\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,dx
\displaystyle \omega=\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{y\sqrt{(\frac{1}{y}-\alpha)(\beta-\frac{1}{y})}}\,dx
\displaystyle =\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{\sqrt{\alpha\beta (\frac{1}{\alpha}-y)(y-\frac{1}{\beta})}}}\,dy
\text{Then we require }Pr(S \cup T | L \cup M)= \dfrac{\frac{1}{3}pq}{_\frac{2}{3}pq+\frac{1}{3}(1-p)(1-q)}=\dfrac{pq}{1-p-q+3pq}}
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1+Tan^2\theta =sec^2\theata
m \mathbf{g}. \mathbf{\gamma}+ \mathbf{T}.\mathbf{\gamma}=0[br]\Rightarrow (m \mathbf{g}+\mathbf{t}).\mathbf{ \gamma}=0[br]\Rightarrow-mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}.\frac{ \mathbf{ \dot{r}}}{|\mathbf{\dot{r}}|}=0 \Rightarrow -mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}. \begin{pmatrix} \dot{r}cos\theta-rsin\theta \\ \dot{r} sin\theta+rcos\theta \end{pmatrix} \frac{1}{\sqrt{\dot{r}^2+r^2}}=0[br]\Rightarrow \dot{r}cos^2\theta-rsin\thetacos\theta+\dot{r} sin\theta+rcos\theta+\dot{r}sin^2\theta+rsin\thetacos\theta=0[br]\Rightarrow \dot{r}(1+sin\theta)+rcos\theta=0[br]\Rightarrow \dot{r}+r\frac{cos\theta}{1+sin\theta}=0
S = (\mathrm{log}_2 e})(1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32...)
S = (\mathrm{log}_2 e})(P + N)
S = (\mathrm{log}_2 e})(P + N) = (\mathrm{log}_2 e})(4/3 - 2/3)
S = 2/3(\mathrm{log}_2 e})
\displaystyle S = 2/3(\mathrm{log}_2 e}) = 2/3(\frac{\mathrm{log}_e e}{\mathrm{log}_e 2}) = 2/3(\frac{1}{\mathrm{log}_e 2})
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