What is the integral from 0 to 1/2 of 5xcos(pix) ?

The integral from 0 to 1/2 of 5xcos(pix) is 5(1/(2pi)-1/(pi^2))

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integral from 0 to 1/2 of 5xcos(pix)

\int_{0}^{\frac{1}{2}} 5 x cos (π x) d x

5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

Decimal

0.28916 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 5 x cos (π x) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{1}{2}} x cos (π x) d x

Apply Integration By Parts

: [\frac{1}{π} (x sin (π x) - π \cdot \int \frac{1}{π} sin (π x) d x)]_{0}^{\frac{1}{2}}

= 5 [\frac{1}{π} (x sin (π x) - π \cdot \int \frac{1}{π} sin (π x) d x)]_{0}^{\frac{1}{2}}

\int \frac{1}{π} sin (π x) d x = - \frac{1}{π ^{2}} cos (π x)

= 5 [\frac{1}{π} (x sin (π x) - π (- \frac{1}{π ^{2}} cos (π x)))]_{0}^{\frac{1}{2}}

Simplify

5 [\frac{1}{π} (x sin (π x) - π (- \frac{1}{π ^{2}} cos (π x)))]_{0}^{\frac{1}{2}} : 5 [\frac{1}{π} (x sin (π x) + \frac{1}{π} cos (π x))]_{0}^{\frac{1}{2}}

= 5 [\frac{1}{π} (x sin (π x) + \frac{1}{π} cos (π x))]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{2 π} - \frac{1}{π ^{2}}

= 5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

\int_{1}^{3} x (x - 1) d x

\int_{0}^{1} c x^{2} + x d x

\int_{0}^{4} π y d y

\int_{0}^{a r c s i n (x)} (cos (t)) d t

\int_{- \frac{π}{2}}^{\frac{π}{2}} I n (cos (x)) d x

What is the integral from 0 to 1/2 of 5xcos(pix) ?
The integral from 0 to 1/2 of 5xcos(pix) is 5(1/(2pi)-1/(pi^2))