What is the integral from 0 to pi/2 of sin^2(2nx) ?

The integral from 0 to pi/2 of sin^2(2nx) is (2pin-sin(2pin))/(8n)

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integral from 0 to pi/2 of sin^2(2nx)

\int_{0}^{\frac{π}{2}} sin^{2} (2 n x) d x

\frac{2 πn - sin ( 2 πn )}{8 n}

Solution steps

\int_{0}^{\frac{π}{2}} sin^{2} (2 n x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} 4 cos^{2} (n x) sin^{2} (n x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (n x) sin^{2} (n x) d x

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 x n )}{8} d x

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 x n) d x

Apply Integration By Parts

: [(1 - cos (4 x n)) x - \int 4 n sin (4 n x) x d x]_{0}^{\frac{π}{2}}

= 4 \cdot \frac{1}{8} [(1 - cos (4 x n)) x - \int 4 n sin (4 n x) x d x]_{0}^{\frac{π}{2}}

\int 4 n sin (4 n x) x d x = - x cos (4 n x) + \frac{1}{4 n} sin (4 n x)

= 4 \cdot \frac{1}{8} [(1 - cos (4 x n)) x - (- x cos (4 n x) + \frac{1}{4 n} sin (4 n x))]_{0}^{\frac{π}{2}}

Simplify

4 \cdot \frac{1}{8} [(1 - cos (4 x n)) x - (- x cos (4 n x) + \frac{1}{4 n} sin (4 n x))]_{0}^{\frac{π}{2}} : \frac{1}{2} [\frac{4 n x - sin ( 4 n x )}{4 n}]_{0}^{\frac{π}{2}}

= \frac{1}{2} [\frac{4 n x - sin ( 4 n x )}{4 n}]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{2 πn - sin ( 2 πn )}{4 n}

= \frac{1}{2} \cdot \frac{2 πn - sin ( 2 πn )}{4 n}

Simplify

= \frac{2 πn - sin ( 2 πn )}{8 n}

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What is the integral from 0 to pi/2 of sin^2(2nx) ?
The integral from 0 to pi/2 of sin^2(2nx) is (2pin-sin(2pin))/(8n)