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

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

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integral from 1/2 to 1 of ycos^2(x)y

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

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

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

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

Simplify

yy : y^{2}

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

Rewrite using trig identities

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

Take the constant out:

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

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

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

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

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

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