What is the integral from 0 to pi/4 of xsin(x)cos(x) ?

The integral from 0 to pi/4 of xsin(x)cos(x) is 1/8

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integral from 0 to pi/4 of xsin(x)cos(x)

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

\frac{1}{8}

Decimal Notation

0.125

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{1}{2} x sin (2 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}^{\frac{π}{4}} x sin (2 x) d x

Apply Integration By Parts

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

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

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{4}

= \frac{1}{2} \cdot \frac{1}{4}

Simplify

= \frac{1}{8}

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What is the integral from 0 to pi/4 of xsin(x)cos(x) ?
The integral from 0 to pi/4 of xsin(x)cos(x) is 1/8