6
$\begingroup$

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.

Please help.

$\endgroup$
5
  • $\begingroup$ You have a mistake in the final equation .. Namely you should a minus sign. $\endgroup$ – Chinny84 Dec 31 '14 at 18:12
  • $\begingroup$ @Chinny84 Where exactly do you think the mistake is? It looks fine to me. $\endgroup$ – David H Dec 31 '14 at 18:17
  • $\begingroup$ @davidh check the edits. But since it has been edited ..my comment is mute. $\endgroup$ – Chinny84 Dec 31 '14 at 18:19
  • $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 31 '14 at 18:37
  • $\begingroup$ Four answers and so far I'm the only person who's up-voted the question. That is often neglected. $\endgroup$ – Michael Hardy Dec 31 '14 at 18:43
10
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

HINT:

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

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

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
  • 1
    $\begingroup$ 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 $\endgroup$ – RE60K Dec 31 '14 at 18:38
0
$\begingroup$

Hint:

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

$\endgroup$
5
  • 2
    $\begingroup$ /begin{pedantry} That's only true if we are assuming (and we are) $0\le x\le \pi$. /end{pedantry} $\endgroup$ – David H Dec 31 '14 at 18:37
  • $\begingroup$ For instance, if $x=\frac{4\pi}{3}$ then the LHS is $\frac{\sqrt{3}-1}{2}$ but the RHS is $-\frac{\sqrt{3}+1}{2}$. $\endgroup$ – David H Dec 31 '14 at 18:46
  • $\begingroup$ @DavidH Yes... so? The integration interval is $\;[0,\pi]\;$, otherwise messing with parts where things are negative can be messy. $\endgroup$ – Timbuc Dec 31 '14 at 18:47
  • $\begingroup$ 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 $\endgroup$ – David H Dec 31 '14 at 18:58
  • $\begingroup$ Oh, I see. It's fine @DavidH , thanks. $\endgroup$ – Timbuc Dec 31 '14 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.