What is the integral of 1/pi xcos(nx) ?

The integral of 1/pi xcos(nx) is 1/(pin^2)(nxsin(nx)+cos(nx))+C

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integral of 1/pi xcos(nx)

\int \frac{1}{π} x cos (n x) d x

\frac{1}{π n ^{2}} (n x sin (n x) + cos (n x)) + C

Solution steps

\int \frac{1}{π} x cos (n x) d x

Take the constant out:

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

= \frac{1}{π} \cdot \int x cos (n x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{n ^{2}} d u

= \frac{1}{π} \cdot \int \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}{π} \cdot \frac{1}{n ^{2}} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= \frac{1}{π} \cdot \frac{1}{n ^{2}} (u sin (u) - \int sin (u) d u)

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

= \frac{1}{π} \cdot \frac{1}{n ^{2}} (u sin (u) - (- cos (u)))

Substitute back

u = n x

= \frac{1}{π} \cdot \frac{1}{n ^{2}} (n x sin (n x) - (- cos (n x)))

Simplify

\frac{1}{π} \cdot \frac{1}{n ^{2}} (n x sin (n x) - (- cos (n x))) : \frac{1}{π n ^{2}} (n x sin (n x) + cos (n x))

= \frac{1}{π n ^{2}} (n x sin (n x) + cos (n x))

Add a constant to the solution

= \frac{1}{π n ^{2}} (n x sin (n x) + cos (n x)) + C

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y = tan (x), at x = - \frac{7 π}{4}

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What is the integral of 1/pi xcos(nx) ?
The integral of 1/pi xcos(nx) is 1/(pin^2)(nxsin(nx)+cos(nx))+C