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

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1. welshfella

maybe expanding it would help x sin x -------- + --------- 1 + cos x 1 + cos x

2. welshfella

- just a thought...

3. anonymous

Let $$u=\tan\dfrac{x}{2}$$, 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{1-u^2}{1+u^2}\end{align*} All this to say that $\int\frac{x+\sin x}{1+\cos x}\,dx=\int\frac{2\arctan u+\dfrac{2u}{1+u^2}}{1+\dfrac{1-u^2}{1+u^2}}\times\frac{2}{1+u^2}\,du$ We're in luck - this simplifies nicely: $4\int\left(\arctan u+\dfrac{u}{1+u^2}\right)\,du$