\(\displaystyle{\int_{{{0}}}^{{{2}\pi}}}\sqrt{{{2}-{2}{\cos{{\left({x}\right)}}}}}{\left.{d}{x}\right.}={2}{\int_{{{0}}}^{{{2}\pi}}}{\sin{{\left({\frac{{{x}}}{{{2}}}}\right)}}}{\left.{d}{x}\right.}\)

Landen Barber

Landen Barber

Answered question

2022-03-29

02π22cos(x)dx=202πsin(x2)dx

Answer & Explanation

microsgopx6z7

microsgopx6z7

Beginner2022-03-30Added 14 answers

HINT: we get
cos(x2+x2)=cos2(x2)sin2(x2)

Aarlenlsi1

Aarlenlsi1

Beginner2022-03-31Added 10 answers

You can recall
1cosx2=sin2x2
If 0x2π, then 0x2π, so sin(x2)0 and therefore
22cosx=2sinx2
So your integral becomes
202πsinx2,dx=2[2cosx2]02π=8
The suggested π2 is definitely wrong

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