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

The integral from 0 to pi/2 of xcos^2(x) is (pi^2-2)/(16)-1/8

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

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

\frac{π ^{2} - 2}{16} - \frac{1}{8}

Decimal

0.36685 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x)]_{0}^{\frac{π}{2}}

= [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x)]_{0}^{\frac{π}{2}}

\int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))

= [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))]_{0}^{\frac{π}{2}} : [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{\frac{π}{2}}

= [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π ^{2} - 2}{16} - \frac{1}{8}

= \frac{π ^{2} - 2}{16} - \frac{1}{8}

\int_{0}^{3} x^{2} - 1 d x

\int_{0}^{t} e^{t} d t

\int_{- 1}^{0} cos (nπ x) d x

\int_{1}^{7} \frac{1}{x ^{2}} d x

\int_{0}^{x} t sin (t) d t

What is the integral from 0 to pi/2 of xcos^2(x) ?
The integral from 0 to pi/2 of xcos^2(x) is (pi^2-2)/(16)-1/8