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

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

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

\int_{0}^{2 π} x cos (2 x) d x

0

Solution steps

\int_{0}^{2 π} x cos (2 x) d x

Apply u-substitution

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

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

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{4 π}

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

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

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

Simplify

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

Compute the boundaries:

0

= \frac{1}{4} \cdot 0

Simplify

= 0

\int_{0}^{1} x^{2} (1 - x)^{5} d x

\int_{0}^{2 π} t cos^{2} (t) d t

\int_{1}^{3} \frac{12 ( ln ( x ) ) ^{5}}{x} d x

\int_{7}^{49} \frac{1}{x ( ln ( x ) ) ^{8}} d x

\int_{0}^{1} (- ln (x)) d x

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