What is the integral from 1/2 to 1 of ycos^2(xy) ?

The integral from 1/2 to 1 of ycos^2(xy) is y/4+(sin(2y)-sin(y))/4

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integral from 1/2 to 1 of ycos^2(xy)

\int_{\frac{1}{2}}^{1} y cos^{2} (x y) d x

\frac{y}{4} + \frac{sin ( 2 y ) - sin ( y )}{4}

Solution steps

\int_{\frac{1}{2}}^{1} y cos^{2} (x y) d x

Take the constant out:

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

= y \cdot \int_{\frac{1}{2}}^{1} cos^{2} (y x) d x

Rewrite using trig identities

= y \cdot \int_{\frac{1}{2}}^{1} \frac{1 + cos ( 2 y x )}{2} d x

Take the constant out:

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

= y \frac{1}{2} \cdot \int_{\frac{1}{2}}^{1} 1 + cos (2 y 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

= y \frac{1}{2} (\int_{\frac{1}{2}}^{1} 1 d x + \int_{\frac{1}{2}}^{1} cos (2 y x) d x)

\int_{\frac{1}{2}}^{1} 1 d x = \frac{1}{2}

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

= y \frac{1}{2} (\frac{1}{2} + \frac{sin ( 2 y ) - sin ( y )}{2 y})

Simplify

y \frac{1}{2} (\frac{1}{2} + \frac{sin ( 2 y ) - sin ( y )}{2 y}) : \frac{y}{4} + \frac{sin ( 2 y ) - sin ( y )}{4}

= \frac{y}{4} + \frac{sin ( 2 y ) - sin ( y )}{4}

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What is the integral from 1/2 to 1 of ycos^2(xy) ?
The integral from 1/2 to 1 of ycos^2(xy) is y/4+(sin(2y)-sin(y))/4