What is the integral from-pi to 0 of-xcos(nx) ?

The integral from-pi to 0 of-xcos(nx) is -(1-(-1)^n)/(n^2)

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integral from-pi to 0 of-xcos(nx)

\int_{- π}^{0} - x cos (n x) d x

- \frac{1 - ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{- π}^{0} - x cos (n x) d x

Take the constant out:

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

= - \int_{- π}^{0} x cos (n x) d x

Apply u-substitution

: \int_{- πn}^{0} \frac{u cos ( u )}{n ^{2}} d u

= - \int_{- πn}^{0} \frac{u cos ( u )}{n ^{2}} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

1 - (- 1)^{n}

= - \frac{1}{n ^{2}} (1 - (- 1)^{n})

Simplify

= - \frac{1 - ( - 1 ) ^{n}}{n ^{2}}

\int_{1}^{2} 2 x^{3} d x

\int_{- 4}^{4} 4^{2} - x^{2} d x

\int_{2}^{2} x d x

\int_{- π}^{π} (π^{2} - x^{2}) d x

\int_{0}^{2 π} 2 cos^{2} (x) d x

What is the integral from-pi to 0 of-xcos(nx) ?
The integral from-pi to 0 of-xcos(nx) is -(1-(-1)^n)/(n^2)