What is the integral from 0 to pi/2 of xcos^3(x) ?

The integral from 0 to pi/2 of xcos^3(x) is pi/3-7/9

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integral from 0 to pi/2 of xcos^3(x)

\int_{0}^{\frac{π}{2}} x cos^{3} (x) d x

\frac{π}{3} - \frac{7}{9}

Decimal

0.26941 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x cos^{3} (x) d x

Apply Integration By Parts

= [x (sin (x) - \frac{1}{3} sin^{3} (x)) - \int \frac{1}{3} (3 sin (x) - sin^{3} (x)) d x]_{0}^{\frac{π}{2}}

\int \frac{1}{3} (3 sin (x) - sin^{3} (x)) d x = \frac{1}{9} (- 6 cos (x) - cos^{3} (x))

= [x (sin (x) - \frac{1}{3} sin^{3} (x)) - \frac{1}{9} (- 6 cos (x) - cos^{3} (x))]_{0}^{\frac{π}{2}}

Simplify

[x (sin (x) - \frac{1}{3} sin^{3} (x)) - \frac{1}{9} (- 6 cos (x) - cos^{3} (x))]_{0}^{\frac{π}{2}} : [\frac{1}{9} (9 x sin (x) - 3 x sin^{3} (x) + 6 cos (x) + cos^{3} (x))]_{0}^{\frac{π}{2}}

= [\frac{1}{9} (9 x sin (x) - 3 x sin^{3} (x) + 6 cos (x) + cos^{3} (x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3} - \frac{7}{9}

= \frac{π}{3} - \frac{7}{9}

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\int_{0}^{2 π} 9 t 2 d t

\int_{- 1}^{2} [(x + 3) - (x^{2} + 1)] d x

\int_{- π}^{π} sin (2 x) cos (2 x) d x

What is the integral from 0 to pi/2 of xcos^3(x) ?
The integral from 0 to pi/2 of xcos^3(x) is pi/3-7/9