What is the integral from 0 to (2pi)/w of cos^2(wt) ?

The integral from 0 to (2pi)/w of cos^2(wt) is pi/w

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integral from 0 to (2pi)/w of cos^2(wt)

\int_{0}^{\frac{2 π}{w}} cos^{2} (wt) d t

\frac{π}{w}

Solution steps

\int_{0}^{\frac{2 π}{w}} cos^{2} (wt) d t

Rewrite using trig identities

= \int_{0}^{\frac{2 π}{w}} \frac{1 + cos ( 2 wt )}{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{2 π}{w}} 1 + cos (2 wt) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{2 π}{w}} 1 d t + \int_{0}^{\frac{2 π}{w}} cos (2 wt) d t)

\int_{0}^{\frac{2 π}{w}} 1 d t = \frac{2 π}{w}

\int_{0}^{\frac{2 π}{w}} cos (2 wt) d t = 0

= \frac{1}{2} (\frac{2 π}{w} + 0)

Simplify

\frac{1}{2} (\frac{2 π}{w} + 0) : \frac{π}{w}

= \frac{π}{w}

\frac{d}{d z} (x^{2} + 2 y z)

f (x) = x^{3} - 12 x + 8

x y^{^{'}} = x^{2} - y^{2} + y

\frac{d y}{d x} + y^{2} = y

\int \frac{1}{x ^{2} - 3 x - 10} d x

What is the integral from 0 to (2pi)/w of cos^2(wt) ?
The integral from 0 to (2pi)/w of cos^2(wt) is pi/w