What is the integral from 0 to 9pi of x(3+sin(x)) ?

The integral from 0 to 9pi of x(3+sin(x)) is (243pi^2+18pi)/2

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integral from 0 to 9pi of x(3+sin(x))

\int_{0}^{9 π} x (3 + sin (x)) d x

\frac{243 π ^{2} + 18 π}{2}

Decimal Notation

1227.43126 \dots

Solution steps

\int_{0}^{9 π} x (3 + sin (x)) d x

Apply Integration By Parts

: [x (3 x - cos (x)) - \int 3 x - cos (x) d x]_{0}^{9 π}

= [x (3 x - cos (x)) - \int 3 x - cos (x) d x]_{0}^{9 π}

\int 3 x - cos (x) d x = \frac{3 x ^{2}}{2} - sin (x)

= [x (3 x - cos (x)) - (\frac{3 x ^{2}}{2} - sin (x))]_{0}^{9 π}

Simplify

[x (3 x - cos (x)) - (\frac{3 x ^{2}}{2} - sin (x))]_{0}^{9 π} : [\frac{1}{2} (3 x^{2} - 2 x cos (x) + 2 sin (x))]_{0}^{9 π}

= [\frac{1}{2} (3 x^{2} - 2 x cos (x) + 2 sin (x))]_{0}^{9 π}

Compute the boundaries:

\frac{243 π ^{2} + 18 π}{2}

= \frac{243 π ^{2} + 18 π}{2}

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\int_{0}^{π} (x + 1) sin (n x) d x

\int_{0}^{3} (2 x - x^{2} + x) d x

\int_{4}^{9} \frac{1}{3 x} d x

What is the integral from 0 to 9pi of x(3+sin(x)) ?
The integral from 0 to 9pi of x(3+sin(x)) is (243pi^2+18pi)/2