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juantheron
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$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{(a+b\cos x)^2}dx$
MHB Definite Integral: $\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{(a+b\cos x)^2}dx$
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The integral $\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{(a+b\cos x)^2}dx$ can be simplified using a substitution that transforms it into an integral from 0 to 1 of a rational function. The function $f(b) = \int_{0}^{\frac{\pi}{2}} \frac{dx}{a + b\cos x}$ is related to the original integral, specifically $\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(a + b\cos x)^{2}} dx = -f'(b)$. By applying the suggested substitution, the integral can be expressed as $f(b) = 2\int_{0}^{1} \frac{1 + t^{2}}{(a+b) + (a-b)t^{2}} dt$. After computation, the result for $f(b)$ is given as $\frac{2}{a+b} + 4 \frac{b}{a-b}\sqrt{\frac{a+b}{a-b}}\tan^{-1}\sqrt{\frac{a-b}{a+b}}$. Deriving this result is left as an exercise for further exploration.
Hello!
I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem.
Given:
##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0##
##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1##
##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0##
I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...
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