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

The integral from pi/2 to pi of xsin^2(x) is (3pi^2-4)/(16)

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integral from pi/2 to pi of xsin^2(x)

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

\frac{3 π ^{2} - 4}{16}

Decimal Notation

1.60055 \dots

Solution steps

\int_{\frac{π}{2}}^{π} x sin^{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)]_{\frac{π}{2}}^{π}

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

\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)))]_{\frac{π}{2}}^{π}

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

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

Compute the boundaries:

\frac{3 π ^{2} - 4}{16}

= \frac{3 π ^{2} - 4}{16}

\frac{d}{d x} (\frac{1 - y}{x - 1})

\int (cos^{2} (x) cos (x)) d x

\int sin^{2} (t) cos^{3} (t) d t

\frac{d}{d x} (\frac{( 1 - x )}{2 x})

y^{^{'}} = 0.19 y + 200

What is the integral from pi/2 to pi of xsin^2(x) ?
The integral from pi/2 to pi of xsin^2(x) is (3pi^2-4)/(16)