What is the integral from 0 to 2pi of |sin(x)cos(x)| ?

The integral from 0 to 2pi of |sin(x)cos(x)| is 2

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integral from 0 to 2pi of |sin(x)cos(x)|

\int_{0}^{2 π} ∣ sin (x) cos (x) ∣ d x

2

Solution steps

\int_{0}^{2 π} ∣ sin (x) cos (x) ∣ d x

Eliminate Absolutes

= \int_{0}^{\frac{π}{2}} sin (x) cos (x) d x + \int_{\frac{π}{2}}^{π} - sin (x) cos (x) d x + \int_{π}^{\frac{3 π}{2}} sin (x) cos (x) d x + \int_{\frac{3 π}{2}}^{2 π} - sin (x) cos (x) d x

\int_{0}^{\frac{π}{2}} sin (x) cos (x) d x = \frac{1}{2}

\int_{\frac{π}{2}}^{π} - sin (x) cos (x) d x = \frac{1}{2}

\int_{π}^{\frac{3 π}{2}} sin (x) cos (x) d x = \frac{1}{2}

\int_{\frac{3 π}{2}}^{2 π} - sin (x) cos (x) d x = \frac{1}{2}

= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}

Simplify

\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} : 2

= 2

\int_{0}^{2} 2 t sin (2 π t) d t

\int_{1}^{2} (x - 1 - (x - 1)^{4}) d x

\int_{0}^{4} \frac{1}{2} x^{2} d x

\int_{- 2}^{2} arctan (x) d x

\int_{0}^{2 - x} \frac{3}{4} x d x

What is the integral from 0 to 2pi of |sin(x)cos(x)| ?
The integral from 0 to 2pi of |sin(x)cos(x)| is 2