What is the integral from 1 to 6 of-3cos((npit)/4) ?

The integral from 1 to 6 of-3cos((npit)/4) is -(3(4sin((3pin)/2)-4sin((pin)/4)))/(pin)

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integral from 1 to 6 of-3cos((npit)/4)

\int_{1}^{6} - 3 cos (\frac{nπ t}{4}) d t

- \frac{3 ( 4 sin ( \frac{3 πn}{2} ) - 4 sin ( \frac{πn}{4} ) )}{πn}

Solution steps

\int_{1}^{6} - 3 cos (\frac{nπ t}{4}) d t

Take the constant out:

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

= - 3 \cdot \int_{1}^{6} cos (\frac{nπ t}{4}) d t

Apply Integration By Parts

: [t cos (\frac{πn t}{4}) - \int - \frac{πn}{4} t sin (\frac{πn t}{4}) d t]_{1}^{6}

= - 3 [t cos (\frac{πn t}{4}) - \int - \frac{πn}{4} t sin (\frac{πn t}{4}) d t]_{1}^{6}

\int - \frac{πn}{4} t sin (\frac{πn t}{4}) d t = t cos (\frac{πn t}{4}) - \frac{4}{πn} sin (\frac{πn t}{4})

= - 3 [t cos (\frac{πn t}{4}) - (t cos (\frac{πn t}{4}) - \frac{4}{πn} sin (\frac{πn t}{4}))]_{1}^{6}

Simplify

= - 3 [\frac{4}{πn} sin (\frac{πn t}{4})]_{1}^{6}

Compute the boundaries:

\frac{4}{πn} sin (\frac{3 πn}{2}) - \frac{4}{πn} sin (\frac{πn}{4})

= - 3 (\frac{4}{πn} sin (\frac{3 πn}{2}) - \frac{4}{πn} sin (\frac{πn}{4}))

Simplify

= - \frac{3 ( 4 sin ( \frac{3 πn}{2} ) - 4 sin ( \frac{πn}{4} ) )}{πn}

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What is the integral from 1 to 6 of-3cos((npit)/4) ?
The integral from 1 to 6 of-3cos((npit)/4) is -(3(4sin((3pin)/2)-4sin((pin)/4)))/(pin)