Question

help please

Answer 1

\[\int\limits_{0}^{\pi/2}\log(sinx)dx\]

Answer 2

@dan815 @nincompoop

Answer 3

@chosenmatt help please

Answer 4

:} instead of me telling you a wrong answer im just gonna tell you i don't know how to solve the question.....sorry :(

Answer 5

fine thanks

Answer 6

@sleepyjess

Answer 7

I haven't done stuff like this yet :/

Answer 8

use P-5 add both the equation and add subtract ln 2 and replace 2x by t in the equation...

Answer 9

hmm
i did not get

Answer 10

still there?

Answer 11

I'm sorry but I do not know how to help you @rvc
http://www.wolframalpha.com/input/?i=integral+of+ln%28sinx%29+from+0+to+pi%2F2

Answer 12

thank u

Answer 13

Not a very easy integral. The indefinite integral is very complicated, but there is a trick for this integral with the given integral limits.

First, we note that
(a) \[ \int_{\alpha}^{\beta}f(x)\, dx =\int_{\alpha}^{\beta} f(\beta +\alpha-x)\, dx\] You can check this easily if you use the general notation \(F(x)\) for the antiderivative.

(b) \[\cos(x)=\sin\left(\frac{\pi}{2}\pm x \right)\]
This should be easily checked by using the property of \(\sin(\theta\pm\phi)=\sin \theta \cos \phi \pm \cos \theta \sin \phi \) on the right side of the equation.

(c) \[ \sin(2x)=2\sin x \cos x\]

(d) Symmetry property: A function is symmetric about the line \(x=a\) if \(f(a+x)=f(a-x)\)
-----
So....
\[ \begin{align} \int_{0}^{\pi/2} \log(\sin x)\, dx &= \int_{0}^{\pi/2} \log\left(\sin \left( \frac{\pi}{2}-x\right)\right)\, dx \qquad \text{by (a)}\\ &= \color{blue}{\int_{0}^{\pi/2} \log(\cos x) \, dx} \qquad \text{by (b)}\ \end{align}\]

This implies:
\[ \begin{align} \color{red}{2}\int_{0}^{\pi/2} \log(\sin x)\, dx &= \int_{0}^{\pi/2} \log(\sin x)\, dx+ \color{blue}{\int_{0}^{\pi/2} \log(\cos x) \, dx}\\
&=\int_{0}^{\pi/2} \left\{\log(\sin x)+\log(\cos x)\right\} \, dx \\
&=\int_{0}^{\pi/2} \log(\sin x \cos x) \, dx \\
&=\int_{0}^{\pi/2} \log\left( \frac{2\, \sin x \cos x}{2}\right) \, dx \\
&= \int_{0}^{\pi/2} \left\{\log(2 \sin x \cos x) -\log(2)\right\} \, dx \\
&= \int_{0}^{\pi/2} \log(\sin(2x)) \, dx - \int_{0}^{\pi/2} \log(2) \, dx \qquad \text{by (c)} \\
&= \int_{0}^{\pi/2} \log(\sin(2x)) \, dx -\frac{\pi}{2}\log 2 \, dx\end{align} \]

Let \(u = 2x \implies \frac{1}{2} du = dx\)
Don't forget to change your bounds of integration:

\[ \begin{align} 2\int_{0}^{\pi/2} \log(\sin x)\, dx&=\int_{0}^{\color{red}{\pi}}\frac{1}{2} \log(\sin u)\, du -\frac{\pi}{2}\log 2 \\
&= \int_{0}^{\color{blue}{\pi/2}}\log(\sin u)\, du - \frac{\pi}{2}\log 2 \qquad \text{by (d)}\end{align} \]
So, \[ 2\int_{0}^{\pi/2} \log(\sin x)\, dx - \int_{0}^{\pi/2} \log(\sin x)\, dx = -\frac{\pi}{2}\log 2 \\
\implies \int_{0}^{\pi/2} \log(\sin x)\, dx = -\frac{\pi}{2}\log 2\]

Answer 14

thank u soooooooo much
but i did not get the D part
cuold u please explain

Answer 15

Sure. Property d) is saying how to determine if a function is symmetric. For example, if you have the parabola \(f(x)=(x-2)^2\), it is symmetric about the line \(x=2\). So it follows that \(f(2-x)=f(2+x)\)

Now, say you are integrating this function from 0 to 4. Since the axis of symmetry occurs at x=2, then the area under the curve from \(x \in [0,2]\) is the same as the area from \(x \in [2, 4]\)



So the symmetry property applies to the function \(\log(\sin x)\) Maybe a picture of the graph helps: http://www.wolframalpha.com/input/?i=log%28sin+x%29

It turns out that the function is symmetric about \(x = \frac{\pi}{2}\), so \(f(\frac{\pi}{2} -x)=f(\frac{\pi}{2}+x)\), so the area (integral) from \(x \in [0, \pi/2]\) is the same value as the integral from \(x \in [\pi/2, \pi]\)

So \(\large \int_0^{\pi} \log(\sin x)\,dx=\int_0^{\pi/2} \log(\sin x)\,dx+\int_{\pi/2}^{\pi} \log(\sin x)\,dx\)

So, \(\large \frac{1}{2} \int_{0}^{\pi}\log(\sin x)\, dx = \int_{0}^{\pi/2} \log(\sin x)\, dx \)
@rvc