What is the integral from 0 to pi/2 of xcos(9x) ?

The integral from 0 to pi/2 of xcos(9x) is (9pi-2)/(162)

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integral from 0 to pi/2 of xcos(9x)

\int_{0}^{\frac{π}{2}} x cos (9 x) d x

\frac{9 π - 2}{162}

Decimal Notation

0.16218 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x cos (9 x) d x

Apply u-substitution

: \int_{0}^{\frac{9 π}{2}} \frac{u cos ( u )}{81} d u

= \int_{0}^{\frac{9 π}{2}} \frac{u cos ( u )}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int_{0}^{\frac{9 π}{2}} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{\frac{9 π}{2}}

= \frac{1}{81} [u sin (u) - \int sin (u) d u]_{0}^{\frac{9 π}{2}}

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

= \frac{1}{81} [u sin (u) - (- cos (u))]_{0}^{\frac{9 π}{2}}

Simplify

= \frac{1}{81} [u sin (u) + cos (u)]_{0}^{\frac{9 π}{2}}

Compute the boundaries:

\frac{9 π}{2} - 1

= \frac{1}{81} (\frac{9 π}{2} - 1)

Simplify

= \frac{9 π - 2}{162}

\int_{0}^{1} \frac{1}{x ^{π}} d x

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{4} d x

\int_{0}^{π} (x + 1) d x

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{3} d x

\int_{1}^{e} (\frac{ln ( u ) + 1}{u}) d u

What is the integral from 0 to pi/2 of xcos(9x) ?
The integral from 0 to pi/2 of xcos(9x) is (9pi-2)/(162)