Definite Integral: int[2pi to 0][sqrt(sinx - cosx + 1)]dx
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Here's a graph of your function. It's a real dilly to integrate.
As TKH said, this type of method is probably your best bet.
This is Simpson's rule using 20 partitions.
If you graph this function, you will find it is not integrable from
This is the graph from
As TKH said, this type of method is probably your best bet.
This is Simpson's rule using 20 partitions.
If you graph this function, you will find it is not integrable from
This is the graph from
Sorry...I was told to use the equation I gave in the previous problem...however it should have been: sqrt((-sinx)^2-(cosx)^2+(1)^2)dt. Then evaluate it from 0 to 2 pi. So in that case could I reduce the sinx^2 + cosx^2 to 1? I would then be left to integrate the sqrt of 2? Thanks and sorry for the mistake!
\(\displaystyle \L\\(sin(x))^{2}-cos^{2}(x)+1=sin^{2}-cos^{2}+1=2-2cos^{2}(x)\)
\(\displaystyle \L\\\sqrt{2-2cos^{2}(x)}=\sqrt{2}|sin(x)|\)
\(\displaystyle \L\\\sqrt{2}\int_{0}^{2{\pi}}|sin(x)|dx\)
The absolute value is difficult to integrate in and of itself. Try setting it up as such:
\(\displaystyle \L\\2\sqrt{2}\int_{0}^{\pi}{sin(x)}dx\)
Do you know what happens if we just used:
\(\displaystyle \L\\{-}\sqrt{2}\int_{0}^{2{\pi}}{sin(x)}dx\ or\ \sqrt{2}\int_{0}^{2{\pi}}{sin(x)}dx?\)
\(\displaystyle \L\\\sqrt{2-2cos^{2}(x)}=\sqrt{2}|sin(x)|\)
\(\displaystyle \L\\\sqrt{2}\int_{0}^{2{\pi}}|sin(x)|dx\)
The absolute value is difficult to integrate in and of itself. Try setting it up as such:
\(\displaystyle \L\\2\sqrt{2}\int_{0}^{\pi}{sin(x)}dx\)
Do you know what happens if we just used:
\(\displaystyle \L\\{-}\sqrt{2}\int_{0}^{2{\pi}}{sin(x)}dx\ or\ \sqrt{2}\int_{0}^{2{\pi}}{sin(x)}dx?\)