How can you evaluate ∫0π2log⁡cos⁡(x)dx ?

Question

How can you evaluate  ∫0π2log⁡cos⁡(x)dx ?

Kirsten Davis · Accepted Answer

Even though there are numerous substitutions, all of the integral variables I employ are x for the purpose of simplicity. since there are many variables that could generate confusion.
I will stand for the integral value. By changing x for $\frac{π}{2} - x$ , we have:
$I = \int_{0}^{\frac{π}{2}} \log \cos (x) d x = \int_{0}^{\frac{π}{2}} \log \sin (x) d x$
And then, we have:
$I = \int_{0}^{\frac{π}{2}} \log (2 \cos (\frac{x}{2}) \sin (\frac{x}{2})) d x$
$= \frac{π}{2} \log 2 + \int_{0}^{\frac{π}{2}} \log \cos (\frac{x}{2}) d x + \int_{0}^{\frac{π}{2}} \log \sin (\frac{x}{2}) d x$
$= \frac{π}{2} + 2 \int_{0}^{\frac{π}{4}} \log \cos (x) d x + 2 \int_{0}^{\frac{π}{4}} + 2 \int_{0}^{\frac{π}{4}} \log \sin (x) d x$
$= \frac{π}{2} \log 2 + I_{1} + I_{2}$
I substitute that in the bottom-second position from the left. $x = \frac{x}{2}$
For $I_{1}$ , use the substitution that $x = \frac{π}{2} - x$ we obtain
$I_{1} = 2 \int_{\frac{π}{4}}^{\frac{π}{2}} \log \sin (x) d x$
It gives that $I_{1} + I_{2} = 2 I$ . So we have
$I = \frac{π}{2} \log 2 + 2 I$
$I = - \frac{π}{2} \log 2$

Shawn Kim · Accepted Answer

∫0π2ln⁡cos⁡xdx=I=∫0π2ln⁡sin⁡xdx  By symmetry we have ln⁡cos⁡x=ln⁡sin⁡x on the interval [0,π.2]. This is true for any even/odd function on this interval, as is an exercise in Demidovich-Problems in Analysis. Thus we have  2I=∫0π2ln⁡cos⁡xdx+∫0π2ln⁡sin⁡xdx=∫0π2ln⁡(sin⁡xcos⁡x)dx=∫0π2ln⁡(12⋅sin⁡(2x))dx  All I used was ln⁡(a⋅b)=ln⁡(a)+ln⁡(b) and 2sin⁡xcos⁡x=sin⁡(2x). Now we split the integral back up to obtain  −∫0π2ln⁡(2)dx+∫0π2ln⁡(sin⁡(2x))dx=2I  But the integral of ln⁡sin⁡u is 2I, thus we have  −πln⁡(2)2+I=2I→I=−πln⁡(2)2

Vasquez · Accepted Answer

We have a well-known identity: ∏k=1n−1sin⁡(πkn)=2n2n and since log⁡sin⁡x is an improperly Riemann-integrable function over (0,π), it follows that: ∫0πlog⁡sin⁡θdθ=limn→+∞πn∑k=1n−1log⁡sin⁡(πkn)=−πlog⁡2 so: ∫0π/2log⁡sin⁡θdθ=∫0π/2log⁡sin⁡θdθ=−π2log⁡2

How can you evaluate \int_0^{\pi/2}\log\cos(x)dx ?

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