What is the integral from 3 to 5 of sin(Inx) ?

The integral from 3 to 5 of sin(Inx) is (-cos(5In)+cos(3In))/(In)

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integral from 3 to 5 of sin(Inx)

\int_{3}^{5} sin (I n x) d x

\frac{- cos ( 5 I n ) + cos ( 3 I n )}{I n}

Solution steps

\int_{3}^{5} sin (I n x) d x

Apply u-substitution

: \int_{I n \cdot 3}^{I n \cdot 5} \frac{sin ( u )}{I n} d u

= \int_{I n \cdot 3}^{I n \cdot 5} \frac{sin ( u )}{I n} d u

Take the constant out:

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

= \frac{1}{I n} \cdot \int_{I n \cdot 3}^{I n \cdot 5} sin (u) d u

Use the common integral:

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

= \frac{1}{I n} [- cos (u)]_{I n \cdot 3}^{I n \cdot 5}

Compute the boundaries:

- cos (5 I n) + cos (3 I n)

= \frac{1}{I n} (- cos (5 I n) + cos (3 I n))

Simplify

= \frac{- cos ( 5 I n ) + cos ( 3 I n )}{I n}

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What is the integral from 3 to 5 of sin(Inx) ?
The integral from 3 to 5 of sin(Inx) is (-cos(5In)+cos(3In))/(In)