Question

Integral of sqrt(sin^2(x)cos^2(x))dx from 0 to 2pi

Answer 1

\[\sqrt{[\sin(x)\cos(x)]^2}=\sin(x)\cos(x) \text{ if } \sin(x)\cos(x)>0 \\ \sqrt{[\sin(x)\cos(x)]^2}=-\sin(x)\cos(x) if \sin(x)\cos(x)<0\]

Answer 2

so looking at the unit circle between 0 and 2pi
both sin and cos are + in the first quadrant so the product of sin and cos is positive
they are opp in the second quadrant so the product will be neg
they are same signed in third quadrant so their product will be pos
they are opp in the fourth so the product will be neg

Answer 3

so what I'm trying to say is this:
\[\int\limits_{0}^{\frac{\pi}{2}}\sin(x)\cos(x) dx -\int\limits_{\frac{\pi}{2}}^{\pi}\sin(x)\cos(x) dx\\+\int\limits_{\pi}^{\frac{3\pi}{2}}\sin(x)\cos(x)dx-\int\limits_{\frac{3\pi}{2}}^{2\pi} \sin(x)\cos(x) dx\]

Answer 4

oh i see

Answer 5

I used the definition of absolute value
since |x|=sqrt(x^2)

Answer 6

now sin(x)cos(x) is an odd function since sin(-x)cos(-x)=-sin(x)cos(x)
So that means we can probably write this as one integral times 4

Answer 7

\[4 \int\limits_{0}^{\frac{\pi}{2}}\sin(x)\cos(x) dx \]

Answer 8

either way though

Answer 9

oh, realizing the odd thing made it a lot simpler, i already did all 4 and combined them :P

Answer 10

lol
it is fine

Answer 11

Either way, it's 2, which is what my teacher and calculator said :)