Question

Integrate

Answer 1

\[\LARGE \int\limits_0^1 \log(\sqrt{1-x}+\sqrt{1+x})dx\]

Answer 2

I substituted \(\Large x=\cos \theta\) and I am on this step somewhere.
\[\LARGE I=\int\limits_0^\frac{\pi}{2}[\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2})] \sin \theta d \theta\]

Answer 3

@Callisto is one who I guess can easily solve it.

Answer 4

\[\LARGE I=\int\limits\limits_0^\frac{\pi}{2}[\log(\ 2 (\sin \frac{\theta}{2}+\frac{\pi}{4})] \sin \theta d \theta\] maybe

Answer 5

hello..............

Answer 6

\[ \int\limits_0^1 \log(\sqrt{1-x}+\sqrt{1+x})dx\]\[ =[x\log(\sqrt{1-x}+\sqrt{1+x})]_0^1 - \int\limits_0^1 xd(\log(\sqrt{1-x}+\sqrt{1+x}))\]\[ =[x\log(\sqrt{1-x}+\sqrt{1+x})]_0^1 - \int\limits_0^1 x(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\times (-\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})dx\]\[ =[x\log(\sqrt{1-x}+\sqrt{1+x})]_0^1 - \int\limits_0^1 x(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\times (-\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})dx\]

Consider\[\int\limits_0^1 x(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\times (-\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}})dx\]\[=\int\limits_0^1 x(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\times (\frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{(1+x)(1-x)}})dx\]\[=\int\limits_0^1 x (\frac{(\sqrt{1-x}-\sqrt{1+x})^2}{-4\sqrt{(1+x)(1-x)}})dx\]\[=\int\limits_0^1 x (\frac{(\sqrt{1-x})^2-2\sqrt{(1-x)(1+x)}+(\sqrt{1+x})^2}{-4\sqrt{(1+x)(1-x)}})dx\]\[=\frac{-1}{4}\int\limits_0^1 x \frac{1-x+1+x}{\sqrt{1-x^2}}-2xdx\]\[=\frac{-1}{4}\int\limits_0^1 \frac{2x}{\sqrt{1-x^2}}-2xdx\]

Answer 7

Eh.. The 3rd and the 4th line are the same :\

Answer 8

hmm..cant we do subsn way

Answer 9

\[ I=\int\limits_0^\frac{\pi}{2}[\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2}))] \sin \theta d \theta\]\[=\int\limits_0^\frac{\pi}{2}[\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2}))] (2\sin\frac{\theta}{2}\cos\frac{\theta}{2}) d \theta\]\[=4\int\limits_0^\frac{\pi}{2}[\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2}))] (\sin\frac{\theta}{2}) \ d (sin\frac{\theta}{2})\]\[=4\int\limits_0^\frac{\sqrt{2}}{2}[\log(\sqrt 2 (u+\sqrt{1-u^2}))] (u) \ d u\]

Answer 10

Not good at using trigo :(
@hartnn Please save us!

Answer 11

omg that is too long. I quit XD

Answer 12

Thanks to some insight from @Callisto's work, I think I've figured it out (in fact, Callisto, you were on the right track, but you simplified your integral incorrectly).

So you want to do integration parts like Callisto originally did; let \(\large u=\log(\sqrt{1-x}+\sqrt{1+x})\) and \(\large \,dv=\,dx\). Then \(\large v=x\) and
\[\large \begin{aligned}\,du &= \frac{1}{\sqrt{1-x}+\sqrt{1+x}} \left(\frac{-1}{2\sqrt{1-x}} +\frac{1}{2\sqrt{1+x}}\right)\,dx \\ &= \frac{\sqrt{1-x}-\sqrt{1+x}}{(1-x)-(1+x)}\left( \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x}\sqrt{1+x}}\right)\,dx\\ &=-\frac{(1-x)-2\sqrt{1-x}\sqrt{1+x}+(1+x) }{4x\sqrt{1-x}\sqrt{1+x}} \,dx\\ &=-\frac{2-2\sqrt{1-x^2} }{4x\sqrt{1-x^2} }\,dx \\ &= -\frac{1-\sqrt{1-x^2}}{2x\sqrt{1-x^2}} \,dx \\ &= -\frac{1}{2} \left(\frac{1}{x\sqrt{1-x^2}} - \frac{1}{x} \right)\,dx\end{aligned}\]
Therefore, if \(\large \displaystyle I=\int_0^1 \log(\sqrt{1-x}+\sqrt{1+x})\,dx\), then we see that
\[\large \begin{aligned} I &= \left.\left[x\log(\sqrt{1-x}+\sqrt{1+x})\right]\right|_0^1 + \frac{1}{2}\int_0^1 x \left(\frac{1}{x\sqrt{1-x^2}} - \frac{1}{x} \right)\,dx \\ &= \left.\left[x\log(\sqrt{1-x}+\sqrt{1+x})\right]\right|_0^1 + \frac{1}{2}\int_0^1 \frac{1}{\sqrt{1-x^2}} - 1\,dx \\ &= \left.\left[x\log(\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2} \arcsin x - \frac{x}{2}\right]\right|_0^1 \\ &= \left(\log(\sqrt{2}) + \frac{1}{2}\arcsin 1 - \frac{1}{2}\right) - \underbrace{\left(0\cdot\log(\sqrt{2}) + \frac{1}{2}\arcsin 0 - 0\right)}_{=0} \\ &= \frac{1}{2}\log 2+\frac{1}{2}\cdot \frac{\pi}{2} - \frac{1}{2}\\ &= \frac{1}{4}\left(2\log 2+\pi - 2\right)\\ &= \frac{1}{4}\left(\log 4 + \pi -2 \right) \end{aligned}\]

Which matches with the answer Mathematica gives me (see attached figure).

Answer 13

The problem is the asker wanted to do it in trigo. substitution.

Answer 14

Sigh.....

Answer 15

use by parts on the other too

Answer 16

I know by part works, despite my poor algebra. But how does trigo sub work in this question?

Answer 17

no trig sub!! just get rid of log by differentiating in by parts.

Answer 18

The asker was asking if he can do it by trigo sub!!!

Answer 19

oh!! i thought they were different integrals.

Answer 20

\[ I=\int\limits_0^\frac{\pi}{2}[\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2}))] \sin \theta d \theta \\ = \log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2})) \int \sin (\theta) - \int \int \sin(\theta) d(\log(\sqrt 2 (\sin \frac{\theta}{2}+\cos \frac{\theta}{2})) )\]
both are same

Answer 21

Why why why?

Answer 22

you are doing subs before and doing IE by parts later.
in original you use IE by parts earlier.

Answer 23

Alright... stiill by parts LOL

Answer 24

\[\begin{align*}
I &= \int_0^1 \log(\sqrt{1-x} +\sqrt{1+x}) dx \\
&=\int_0^1 \frac 1 2 \log (2 + 2 \sqrt{1-x^2})dx \\
&= \frac 1 2 \int_0^1 (\log 2 + \log (1 + \sqrt {1-x^2} ))dx \\
&= \frac 1 2 \log 2 + \frac 1 2 \int_0^1 \log (1 + \sqrt{1-x^2})dx\\
&= \frac 1 2 \log 2 + \frac 1 2 \int_0^{\frac \pi 2} \log \left( 2 \cos \left(\frac \theta 2\right)\right) \cos(\theta )d\theta \\
&= \frac 1 2 \log 2 + \frac {\log 2} 2 + \frac 1 2 \int_0^{\frac \pi 2} \log \left( \cos \left(\frac \theta 2\right)\right) \cos(\theta )d\theta
\end{align*} \]
Still I can't see how IE by parts can be avoided.

Answer 25

@hartnn

Answer 26

http://www.wolframalpha.com/input/?i=integral_0%5E1+log%28%28%281-x%29%5E1%2F2%29%2B%281%2Bx%29%5E1%2F2%29+dx&a=*FunClash.log-_*Log10-

Answer 27

or this depending on base..
http://www.wolframalpha.com/input/?i=integral_0%5E1+log%28%28%281-x%29%5E1%2F2%29%2B%281%2Bx%29%5E1%2F2%29+dx&a=*FunClash.log-_*Log-