Show that with limits pi/3 – pi/2, ∫ sin x [(1 + cos x) / (1 - cos x)] dx = ln 4 -1/2. How can I do this?
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Originally Posted by geton Show that with limits pi/3 – pi/2, ∫ sin x [(1 + cos x) / (1 - cos x)] dx = ln 4 -1/2. How can I do this? Substitute $\displaystyle u = \cos x$ and $\displaystyle dx = - \frac{du}{\sin x}$. Make sure to substitute for the integral terminals as well.
Originally Posted by mr fantastic Substitute $\displaystyle u = \cos x$ and $\displaystyle dx = - \frac{du}{\sin x}$. Make sure to substitute for the integral terminals as well. Thank you so much.
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