What is the integral from pi/3 to pi of xcos(2x) ?

The integral from pi/3 to pi of xcos(2x) is (3sqrt(3)-2pi)/(8sqrt(3))

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integral from pi/3 to pi of xcos(2x)

\int_{\frac{π}{3}}^{π} x cos (2 x) d x

\frac{3 3 - 2 π}{8 3}

Decimal Notation

- 0.07844 \dots

Solution steps

\int_{\frac{π}{3}}^{π} x cos (2 x) d x

Apply u-substitution

: \int_{\frac{2 π}{3}}^{2 π} \frac{u cos ( u )}{4} d u

= \int_{\frac{2 π}{3}}^{2 π} \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_{\frac{2 π}{3}}^{2 π} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{\frac{2 π}{3}}^{2 π}

= \frac{1}{4} [u sin (u) - \int sin (u) d u]_{\frac{2 π}{3}}^{2 π}

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

= \frac{1}{4} [u sin (u) - (- cos (u))]_{\frac{2 π}{3}}^{2 π}

Simplify

= \frac{1}{4} [u sin (u) + cos (u)]_{\frac{2 π}{3}}^{2 π}

Compute the boundaries:

\frac{3}{2} - \frac{π}{3}

= \frac{1}{4} (\frac{3}{2} - \frac{π}{3})

Simplify

= \frac{3 3 - 2 π}{8 3}

\int_{0}^{1} \frac{1}{2 + cos ( π x )} d x

\int_{- π}^{0} (t + \frac{π}{2}) cos (n t) d t

\int_{0}^{2} \frac{3 x}{4} d x

\int_{- 4}^{- 1} (x^{3} + 2 x^{2} + 1) d x

\int_{- 3}^{5} [2 y - (y^{2} - 15)] d y

What is the integral from pi/3 to pi of xcos(2x) ?
The integral from pi/3 to pi of xcos(2x) is (3sqrt(3)-2pi)/(8sqrt(3))