What is the integral from 9 to 15 of 2cos((pix)/6) ?

The integral from 9 to 15 of 2cos((pix)/6) is (24)/pi

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integral from 9 to 15 of 2cos((pix)/6)

\int_{9}^{15} 2 cos (\frac{π x}{6}) d x

\frac{24}{π}

Decimal Notation

7.63943 \dots

Solution steps

\int_{9}^{15} 2 cos (\frac{π x}{6}) d x

Take the constant out:

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

= 2 \cdot \int_{9}^{15} cos (\frac{π x}{6}) d x

Apply u-substitution

: \int_{\frac{3 π}{2}}^{\frac{5 π}{2}} cos (u) \frac{6}{π} d u

= 2 \cdot \int_{\frac{3 π}{2}}^{\frac{5 π}{2}} cos (u) \frac{6}{π} d u

Take the constant out:

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

= 2 \cdot \frac{6}{π} \cdot \int_{\frac{3 π}{2}}^{\frac{5 π}{2}} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{6}{π} [sin (u)]_{\frac{3 π}{2}}^{\frac{5 π}{2}}

Simplify

2 \cdot \frac{6}{π} [sin (u)]_{\frac{3 π}{2}}^{\frac{5 π}{2}} : \frac{12}{π} [sin (u)]_{\frac{3 π}{2}}^{\frac{5 π}{2}}

= \frac{12}{π} [sin (u)]_{\frac{3 π}{2}}^{\frac{5 π}{2}}

Compute the boundaries:

2

= \frac{12}{π} \cdot 2

Simplify

= \frac{24}{π}

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What is the integral from 9 to 15 of 2cos((pix)/6) ?
The integral from 9 to 15 of 2cos((pix)/6) is (24)/pi