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

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

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integral from 0 to x of xcos(ax)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

\frac{d y}{d x} + y = sin (x)

\int_{- \frac{π}{4}}^{0} 2 sec^{3} (x) d x

\frac{d}{d x} (1 - \frac{x}{2} - ln (2 + 9 x - x^{2}))

\int e^{x} sin (x) sin (3 x) d x

\frac{d}{d x} (x^{\frac{1}{x ^{2}}})

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