What is the integral from 0 to pi of tcos(2t) ?

The integral from 0 to pi of tcos(2t) is 0

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integral from 0 to pi of tcos(2t)

\int_{0}^{π} t cos (2 t) d t

0

Solution steps

\int_{0}^{π} t cos (2 t) d t

Apply u-substitution

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

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

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

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

Simplify

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

Compute the boundaries:

0

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

Simplify

= 0

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What is the integral from 0 to pi of tcos(2t) ?
The integral from 0 to pi of tcos(2t) is 0