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

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

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

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

0

Solution steps

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

Apply u-substitution

: \int_{0}^{2 π} \frac{u cos ( u )}{16} d u

= \int_{0}^{2 π} \frac{u cos ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{0}^{2 π} u cos (u) d u

Apply Integration By Parts

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

= \frac{1}{16} [u sin (u) - \int sin (u) d u]_{0}^{2 π}

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

= \frac{1}{16} [u sin (u) - (- cos (u))]_{0}^{2 π}

Simplify

= \frac{1}{16} [u sin (u) + cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{16} \cdot 0

Simplify

= 0

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