What is the integral from 0 to 2 of-g*cos(n*pi*g) ?

The integral from 0 to 2 of-g*cos(n*pi*g) is -(2sin(2pin))/(pin)

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integral from 0 to 2 of-gcos(npi*g)

\int_{0}^{2} - g \cdot cos (n \cdot π \cdot g) d g

- \frac{2 sin ( 2 πn )}{πn}

Solution steps

\int_{0}^{2} - g cos (nπ g) d g

Take the constant out:

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

= - \int_{0}^{2} g cos (nπ g) d g

Apply u-substitution

: \int_{0}^{n \cdot 2 π} \frac{u cos ( u )}{π ^{2} n ^{2}} d u

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

2 πn sin (2 πn)

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

Simplify

= - \frac{2 sin ( 2 πn )}{πn}

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What is the integral from 0 to 2 of-g*cos(n*pi*g) ?
The integral from 0 to 2 of-g*cos(n*pi*g) is -(2sin(2pin))/(pin)