Anyone know how to do this integral, I am confused? (1 Viewer)

v.tex · Monday at 5:20 PM

Has it got something to do with even/odd functions- symmetry?

Also does it involve partial fractions

cheesynooby · Monday at 7:05 PM

isn't the bottom just a perfect square, and then it's just a standard integral

since the integrand becomes "sin(2π/5) [x + cos(2π/5)]^-2"

tywebb · Monday at 7:06 PM

$\frac{\pi}{2}$

i put into wolframalpha and made it show steps

v.tex · Monday at 8:28 PM

tywebb said:
$\frac{\pi}{2}$

i put into wolframalpha and made it show steps

View attachment 47419
View attachment 47420
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View attachment 47422

Do you have to put it in surd form, can you leave the trig?

tywebb · Monday at 8:30 PM

v.tex said:
Do you have to put it in surd form, can you leave the trig?

i don’t think that matters. answer will be the same.

tywebb · Tuesday at 7:43 AM

i can do it better than the wolframalpha utilising these results:

$\textstyle\int\frac{1}{(x+a)^2+b^2}\ \!dx=\frac{1}{b}\tan^{-1}\frac{a+x}{b}+C$

$\textstyle\csc x+\cot x=\cot\frac{x}{2}\text{ and }\csc x-\cot x=\tan\frac{x}{2}$

$\textstyle\text{For }0<x<\frac{\pi}{2},\tan^{-1}\tan x=x\text{ and }\tan^{-1}\cot x=\frac{\pi}{2}-x$

and now

$\begin{aligned}\textstyle\int_{-1}^1\frac{\sin\frac{2\pi}{5}}{x^2+2x\cos\frac{2\pi}{5}+1}\ \!dx&=\textstyle\int_{-1}^1\frac{\sin\frac{2\pi}{5}}{x^2+2x\cos\frac{2\pi}{5}+\cos^2\frac{2\pi}{5}+\sin^2\frac{2\pi}{5}}\ \!dx\\ &=\textstyle\sin\frac{2\pi}{5}\int_{-1}^1\frac{1}{(x+\cos\frac{2\pi}{5})^2+\sin^2\frac{2\pi}{5}}\ \!dx\\ &=\textstyle\left[\tan^{-1}\frac{\cos\frac{2\pi}{5}+x}{\sin\frac{2\pi}{5}}\right]_{-1}^1\\ &=\tan^{-1}\frac{\cos\frac{2\pi}{5}+1}{\sin\frac{2\pi}{5}}-\tan^{-1}\frac{\cos\frac{2\pi}{5}-1}{\sin\frac{2\pi}{5}}\\ &=\textstyle\tan^{-1}(\csc\frac{2\pi}{5}+\cot\frac{2\pi}{5})+\tan^{-1}(\csc\frac{2\pi}{5}-\cot\frac{2\pi}{5})\\ &=\textstyle\tan^{-1}\cot\frac{\pi}{5}+\tan^{-1}\tan\frac{\pi}{5}\\ &=\textstyle\frac{\pi}{2}-\frac{\pi}{5}+\frac{\pi}{5}\\ &=\textstyle\frac{\pi}{2}\end{aligned}$

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Anyone know how to do this integral, I am confused? (1 Viewer)

v.tex

Well-Known Member

cheesynooby

Well-Known Member

tywebb

dangerman

v.tex

Well-Known Member

tywebb

dangerman

tywebb

dangerman

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