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Find this integral: $$\int\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}\mathrm dx$$

My try: since $$1+\sin{x}+\cos{x}=2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}$$ $$\cot{x}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}}$$ so $$\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}\left(2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}\right)}$$ then I fell very ugly.Thank you

jimjim
  • 9,675

3 Answers3

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Now divide the numerator & the denominator by $\displaystyle \sec^2\frac x2=1+\tan^2\frac x2$

So, we have $I=\displaystyle\int\frac{1-\tan^2\frac x2}{2\tan\frac x2(2+2\tan\frac x2)}\sec^2\frac x2dx$

Putting $\displaystyle \tan\frac x2=u$

$$I=\int\frac{1-u^2}{2u(1+u)}du=\frac12\int\frac1u du-\frac12\int du$$ assuming $1+u\ne0$

lab bhattacharjee
  • 274,582
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Note that \begin{align} \frac{\cot{x}}{1+\sin{x}+\cos{x}} =& \ \frac{\cos{x}(1+\sin{x}+\cos{x})}{\sin x(1+\sin{x}+\cos{x})^2}\\ =& \ \frac{(1+\sin x )(1+\cos x)-\sin x(1+\sin x)}{2\sin x(1+\sin x )(1+\cos x)} \end{align} Thus \begin{align} \int\frac{\cot{x}}{1+\sin{x}+\cos{x}}\ dx =& \ \frac12 \int \left(\frac1{\sin x }- \frac1{1+\cos x}\right)dx\\ =& \ -\frac12\coth^{-1}(\sec x)-\frac12( \csc x-\cot x) \end{align}

Quanto
  • 97,352
2

Let $$I=\int \frac{\cot x}{1+\sin x+\cos x} \operatorname{d}x$$ Substitute $t = \tan\left(\frac{x}{2}\right)$ and $\operatorname{d}t = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \operatorname{d}x$, and transform the integrand using the substitutions $\sin x = \frac{2 t}{t^2+1}, \cos x = \frac{1-t^2}{t^2+1}$ and $\operatorname{d}x = \frac{2 }{t^2+1}\operatorname{d}t$: $$I= \int -\frac{t^2-1}{t(t^2+1)\left(1+\frac{2 t}{t^2+1}+\frac{1-t^2}{t^2+1}\right)} \operatorname{d}t= \int \frac{1}{2}\left(\frac{1}{t}-1\right) \operatorname{d}t=\frac{1}{2}(\ln t+t)+C$$

Substitute back for $t = \tan\left(\tfrac{x}{2}\right)$ and finally $$ I = \frac{1}{2}\left[\ln\left(\tan\left(\tfrac{x}{2}\right)\right)-\tan\left(\tfrac{x}{2}\right)\right]+C. $$

alexjo
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