$\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx$

I ran into this integral and I'm trying to solve it.

$$I=\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx$$

Well, this has something to do with symmetries.

At first I applied the sub $u=23\pi+71\dfrac{\pi}{2}-x$ but I didn't like the result that this sub gave me, so I didn't dwelve further. Then I thought of applying log properties so that the integral was rewritten as: \begin{aligned} &\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx\\ &=\int_{23\pi}^{71\pi/2}\left[ \left ( 1+\cos x \right )\ln \left ( 1+\sin x \right )- \ln \left ( 1+\cos x \right )\right ] \,dx\\ &= \int_{23\pi}^{71\pi/2}\left ( 1+\cos x \right )\ln \left ( 1+\sin x \right )dx-\int_{23\pi}^{71\pi/2}\ln \left ( 1+\cos x \right )\,dx \end{aligned}

then substite $u=1+\sin x$ or $u=1+\cos x$ change the limits of integration, but then again I don't get much help out of it, because the limits are bizzare (at least to me). If the limits were OK, then I would apply IBP to both integrals (definitely not a good way to go around), however I have not also dwelved in it further. I also didn't check if by splitting the integral apart I get into improper ones, or one of them is not defined on the given interval.

Any ideas?

Edit: Some typo correction was done!

• see last form it should be $(1+\cos x)\ln(1+\sin x)$ – RE60K Oct 2 '14 at 11:03
• Oops... $\LaTeX$ demon... ! Yes, that should be.. But then it should not affect my procedure on that case... I would sub something of the two... (if that eventually worked) – Tolaso Oct 2 '14 at 11:06
• What did you mean by $71\pi/2$ and $71\dfrac{\pi}{2}$? I don't get it those two notations. Is it $\dfrac{71\pi}{2}$ or $71\dfrac{1}{2}\pi$? – Anastasiya-Romanova 秀 Oct 2 '14 at 11:16
• @Tolaso $71\dfrac{\pi}{2}\neq\dfrac{71\pi}{2}$ – Anastasiya-Romanova 秀 Oct 2 '14 at 11:28
• @robjohn Notation $71\dfrac{\pi}{2}$ is really weird to me and I can assure you it's not allowed to use it in high school, but I don't know in college. If $71\dfrac{\pi}{2}$ means a product, I think it will be nice if we write it as $71\cdot\dfrac{\pi}{2}$. In my opinion mixed fractions is okay as long as the all numbers are rational like $71\dfrac{1}{2}$. – Anastasiya-Romanova 秀 Oct 2 '14 at 13:19

Some hints:

• Since $\sin$ and $\cos$ are periodic with period $T=2\pi$, $\int_T^{a+T} f(\sin x,\cos x)dx=\int_0^af(\sin x,\cos x)dx$
• $\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ which is particulary useful with trigonometric functions.
• $\int_0^2a f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$

Part of your Integral: $$I=\int_{23\pi}^{71\pi/2}\ln(1+\cos x)dx\\I=\int_0^{25\pi/2}\ln(1+\cos x)dx\\I=12\int_0^{\pi}\ln(1+\cos x)dx+\int_0^{\pi/2}\ln(1+\cos x)dx\\I=12\int_0^{\pi}\ln(1-\cos x)dx+\int_0^{\pi/2}\ln(1+\sin x)dx\\2I=12\int_0^{\pi}\ln((1+\cos x)(1-\cos x))+2\int_0^{\pi/2}\ln(1+\cos x)dx\\ 2I=24\int_0^{\pi}\ln(\sin x)+2\int_0^{\pi/2}\ln(1+\cos x)dx$$

Intermediate: $$\int_0^{\pi}\ln(\sin x)\\=\int_0^{\pi/2}\ln(\sin x)+\ln(\sin (\pi/2-x))\\=\int_0^{\pi/2}\ln(\sin 2x)-\ln(2)\\=0.5\int_0^{\pi}\sin(x)dx-\pi\ln(2)/2$$

So, $$I=12(-\pi\ln(2))+2\int_0^{\pi/2}\ln(1+\cos x)$$ Now your integral is(let it be J): $$J=I+\int_0^{25\pi/2}(1+\cos x)\ln(1+\sin x)\\ \small K(\text{let})=J+\pi\ln(2)=2\int_0^{\pi/2}\ln(1+\cos x)+12\int_0^{\pi}(1+\cos x)\ln(1+\sin x)+\int_0^{\pi/2}(1+\cos x)\ln(1+\sin x)\\=\ldots\\$$ Hereafter make it in form $\int(P(x))\ln(1+\cos x)$. I hope you can do from here?

• That will do... Thank you! – Tolaso Oct 2 '14 at 11:22
• @Tolaso edited.. – RE60K Oct 2 '14 at 11:23
• Thanks... Something I did not count on. – Tolaso Oct 2 '14 at 11:27
• @Tolaso Reedited – RE60K Oct 2 '14 at 11:49

Note that $$\int\left[\vphantom{\sum}\log(1+\sin(x))-\log(1+\cos(x))\right]\mathrm{d}x\tag{1}$$ is $0$ over quadrants $1$ and $3$ and is opposite over quadrants $2$ and $4$. Thus, the integral is $0$ over the whole circle.

Furthermore, \begin{align} \int\cos(x)\log(1+\sin(x))\,\mathrm{d}x &=\int\log(1+\sin(x))\,\mathrm{d}(1+\sin(x))\\ &=(1+\sin(x))\log(1+\sin(x))-\sin(x)+C\tag{2} \end{align} Thus, this integral is $0$ over the whole circle as well.

Note that your integral is the sum of $(1)$ and $(2)$

The domain of integration is $[23\pi,35\pi]\cup[35\pi,71\pi/2]$. $[23\pi,35\pi]$ consists of $6$ full circles and $[35\pi,71\pi/2]$ is quadrant $3$. Thus, we only need compute $(2)$ over quadrant $3$, and that is $1$.

Therefore, $$\int_{23\pi}^{71\pi/2}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x=1\tag{3}$$