What is the integral from 0 to pi/4 of xcos(2x) ?

The integral from 0 to pi/4 of xcos(2x) is (pi-2)/8

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integral from 0 to pi/4 of xcos(2x)

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

\frac{π - 2}{8}

Decimal

0.14269 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{π}{2}} \frac{u cos ( u )}{4} d u

= \int_{0}^{\frac{π}{2}} \frac{u cos ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{\frac{π}{2}} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{\frac{π}{2}}

= \frac{1}{4} [u sin (u) - \int sin (u) d u]_{0}^{\frac{π}{2}}

\int sin (u) d u = - cos (u)

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

Simplify

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

Compute the boundaries:

\frac{π}{2} - 1

= \frac{1}{4} (\frac{π}{2} - 1)

Simplify

= \frac{π - 2}{8}

y^{^{'}} x = y^{3}

\int_{0}^{2 π} \int_{0}^{a} p d p d t

\int \frac{1}{( x ^{2} + 1 ) arctan ( x )} d x

\int x^{2} \cdot cos (2 x) d x

\frac{d}{d x} (- \frac{2}{cos ( 2 x )} + \frac{8}{3})

What is the integral from 0 to pi/4 of xcos(2x) ?
The integral from 0 to pi/4 of xcos(2x) is (pi-2)/8