What is the integral from 0 to 9pi of x(5+sin(x)) ?

The integral from 0 to 9pi of x(5+sin(x)) is (405pi^2+18pi)/2

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integral from 0 to 9pi of x(5+sin(x))

\int_{0}^{9 π} x (5 + sin (x)) d x

\frac{405 π ^{2} + 18 π}{2}

Decimal Notation

2026.86922 \dots

Solution steps

\int_{0}^{9 π} x (5 + sin (x)) d x

Apply Integration By Parts

: [x (5 x - cos (x)) - \int 5 x - cos (x) d x]_{0}^{9 π}

= [x (5 x - cos (x)) - \int 5 x - cos (x) d x]_{0}^{9 π}

\int 5 x - cos (x) d x = \frac{5 x ^{2}}{2} - sin (x)

= [x (5 x - cos (x)) - (\frac{5 x ^{2}}{2} - sin (x))]_{0}^{9 π}

Simplify

[x (5 x - cos (x)) - (\frac{5 x ^{2}}{2} - sin (x))]_{0}^{9 π} : [\frac{1}{2} (5 x^{2} - 2 x cos (x) + 2 sin (x))]_{0}^{9 π}

= [\frac{1}{2} (5 x^{2} - 2 x cos (x) + 2 sin (x))]_{0}^{9 π}

Compute the boundaries:

\frac{405 π ^{2} + 18 π}{2}

= \frac{405 π ^{2} + 18 π}{2}

\int_{- 2}^{2} 2 (4 - x^{2}) d x

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\int_{- 1}^{6} [5 y - (y^{2} - 6)] d y

\int_{2}^{7} \frac{3}{x - 2} d x

\int_{y}^{1} e^{y^{2}} d y

What is the integral from 0 to 9pi of x(5+sin(x)) ?
The integral from 0 to 9pi of x(5+sin(x)) is (405pi^2+18pi)/2