What is the integral from 0 to 2pi of sin^2(7x) ?

The integral from 0 to 2pi of sin^2(7x) is pi

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integral from 0 to 2pi of sin^2(7x)

\int_{0}^{2 π} sin^{2} (7 x) d x

π

Solution steps

\int_{0}^{2 π} sin^{2} (7 x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (1 - cos (14 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 π} 1 - cos (14 x) d x

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}^{2 π} 1 d x - \int_{0}^{2 π} cos (14 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (14 x) d x = 0

= \frac{1}{2} (2 π - 0)

Simplify

\frac{1}{2} (2 π - 0) : π

= π

\int_{0}^{s i n (x)} 4 t d t

\int sin^{5} (x co) s^{5} x d x

7 x (x + 5 y) (\frac{d y}{d x}) = 7 y (x - 5 y)

\frac{1}{s ( s ^{2} + 36 )}

(4 x + 4 y)^{3} = 64 x^{3} + 64 y^{3}, (- 1, 1)

What is the integral from 0 to 2pi of sin^2(7x) ?
The integral from 0 to 2pi of sin^2(7x) is pi