What is the integral from 0 to 1 of xcos(ax) ?

The integral from 0 to 1 of xcos(ax) is (asin(a)+cos(a)-1)/(a^2)

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integral from 0 to 1 of xcos(ax)

\int_{0}^{1} x cos (a x) d x

\frac{a sin ( a ) + cos ( a ) - 1}{a ^{2}}

Solution steps

\int_{0}^{1} x cos (a x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

a sin (a) + cos (a) - 1

= \frac{1}{a ^{2}} (a sin (a) + cos (a) - 1)

Simplify

= \frac{a sin ( a ) + cos ( a ) - 1}{a ^{2}}

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What is the integral from 0 to 1 of xcos(ax) ?
The integral from 0 to 1 of xcos(ax) is (asin(a)+cos(a)-1)/(a^2)