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anonymous
 one year ago
Integrate (x+sinx)/(1+cosx)
anonymous
 one year ago
Integrate (x+sinx)/(1+cosx)

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1maybe expanding it would help x sin x  +  1 + cos x 1 + cos x

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1 just a thought...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let \(u=\tan\dfrac{x}{2}\), or \(x=2\arctan u\), so that \(dx=\dfrac{2}{1+u^2}\,du\). Since \(\tan\dfrac{x}{2}=u\), it follows that \(\sin\dfrac{x}{2}=\dfrac{u}{\sqrt{1+u^2}}\) and \(\cos\dfrac{x}{2}=\dfrac{1}{\sqrt{1+u^2}}\). Recall some identities: \[\begin{align*}\sin x&=\sin\left(2\times\frac{x}{2}\right)\\[1ex]&=2\sin\frac{x}{2}\cos\frac{x}{2}\\[1ex]&=\frac{2u}{1+u^2}\\[2ex] \cos x&=\cos\left(2\times\frac{x}{2}\right)\\[1ex]&=\cos^2\frac{x}{2}\sin^2\frac{x}{2}\\[1ex]&=\frac{1u^2}{1+u^2}\end{align*}\] All this to say that \[\int\frac{x+\sin x}{1+\cos x}\,dx=\int\frac{2\arctan u+\dfrac{2u}{1+u^2}}{1+\dfrac{1u^2}{1+u^2}}\times\frac{2}{1+u^2}\,du\] We're in luck  this simplifies nicely: \[4\int\left(\arctan u+\dfrac{u}{1+u^2}\right)\,du\]
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