# Integrals of Trigonometry functions

Show that

$$\int^\pi_0 \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx =\int^\pi_0 \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx$$

I tried to use to check if

$$\int^\pi_0\left(\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}} - \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\right)\,dx=0$$

but it didn't turn out well.

• You have a mistake in the final equation .. Namely you should a minus sign. – Chinny84 Dec 31 '14 at 18:12
• @Chinny84 Where exactly do you think the mistake is? It looks fine to me. – David H Dec 31 '14 at 18:17
• @davidh check the edits. But since it has been edited ..my comment is mute. – Chinny84 Dec 31 '14 at 18:19
• The point would be that the integral of the first function from $0$ to $\pi/2$ is the same as the integral of the second function from $\pi/2$ to $\pi$ and vice-versa. See details in my answer below. ${}\qquad{}$ – Michael Hardy Dec 31 '14 at 18:37
• Four answers and so far I'm the only person who's up-voted the question. That is often neglected. – Michael Hardy Dec 31 '14 at 18:43

HINT: Substitute $u=\pi-x$ in the first integral.

Your idea actually works; you just have to pursue it. Abbreviating $\cos x$ to $c$ (and $\sin x$ to $s$), we have

$${\sqrt{1+c}-\sqrt{1-c}\over\sqrt{1+c}+\sqrt{1-c}}={(\sqrt{1+c}-\sqrt{1-c})^2\over(1+c)-(1-c)}={1+c-2\sqrt{1-c^2}+1-c\over2c}={1-s\over c}={c\over1+s}$$

(NB: $\sqrt{1-c^2}=s$ has the correct sign, since $s=\sin x\ge0$ for $0\le x\le\pi$.) The substitution $u=1+\sin x$, $du=\cos x\ dx$ gives

$$\int_0^\pi{\sqrt{1+\cos x}-\sqrt{1-\cos x}\over\sqrt{1+\cos x}+\sqrt{1-\cos x}}dx=\int_0^\pi{\cos x\ dx\over1+\sin x}=\int_1^1{du\over u}=0$$

HINT:

Use $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ which can be derived using $a+b-x=y$

and $\cos(\pi-x)=-\cos x$

The point would be that the integral of the first function from $0$ to $\pi/2$ is the same as the integral of the second function from $\pi/2$ to $\pi$ and vice-versa. \begin{align} & \int_{x=0}^{x=\pi/2} \Big(\text{some function of } \cos x\Big)\,dx \\[6pt] = {} & \int_{u=\pi}^{u=\pi/2} \Big(\text{the same function of }\cos (\pi-u)\Big)\,(-du) \\[6pt] = {} & \int_{u=\pi}^{u=\pi/2} \Big(\text{the same function of }(-\cos u)\Big)\,(-du) \\[6pt] = {} & \int_{\pi/2}^\pi \Big(\text{the same function of }(-\cos u)\Big)\,du \\[6pt] = {} & \int_{\pi/2}^\pi \Big(\text{the same function of }(-\cos x)\Big)\,dx \end{align}

• very unpleasing to eyes answer can you remove the dots and make it a bit eye candy, sorry If I am being rude or getting personnel, In any case the decision is yours – RE60K Dec 31 '14 at 18:38

Hint:

$$\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}=\frac1{2\cos x}+\frac12-\frac12\tan x$$

• /begin{pedantry} That's only true if we are assuming (and we are) $0\le x\le \pi$. /end{pedantry} – David H Dec 31 '14 at 18:37
• For instance, if $x=\frac{4\pi}{3}$ then the LHS is $\frac{\sqrt{3}-1}{2}$ but the RHS is $-\frac{\sqrt{3}+1}{2}$. – David H Dec 31 '14 at 18:46
• @DavidH Yes... so? The integration interval is $\;[0,\pi]\;$, otherwise messing with parts where things are negative can be messy. – Timbuc Dec 31 '14 at 18:47
• I was merely pointing out that while the two sides happen to coincide on the integration interval, they aren't equal in general. If you think I'm being overly pedantic here, I did try to warn you. =p – David H Dec 31 '14 at 18:58
• Oh, I see. It's fine @DavidH , thanks. – Timbuc Dec 31 '14 at 23:11