What is the integral from 0 to pi of cos(x+nx) ?

The integral from 0 to pi of cos(x+nx) is 0

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integral from 0 to pi of cos(x+nx)

\int_{0}^{π} cos (x + n x) d x

0

Solution steps

\int_{0}^{π} cos (x + n x) d x

Apply u-substitution

: \int_{0}^{π + nπ} \frac{cos ( u )}{1 + n} d u

= \int_{0}^{π + nπ} \frac{cos ( u )}{1 + n} d u

Take the constant out:

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

= \frac{1}{1 + n} \cdot \int_{0}^{π + nπ} cos (u) d u

Use the common integral:

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

= \frac{1}{1 + n} [sin (u)]_{0}^{π + nπ}

Compute the boundaries:

0

= \frac{1}{1 + n} \cdot 0

Simplify

= 0

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What is the integral from 0 to pi of cos(x+nx) ?
The integral from 0 to pi of cos(x+nx) is 0