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

The integral from 0 to pi/2 of x^2cos^2(x) is (pi^3-6pi)/(48)

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

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

\frac{π ^{3} - 6 π}{48}

Decimal

0.25326 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x ^{3} sin ( 2 x )}{3} d x = \frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{3} (x^{3} cos^{2} (x) - 3 (\frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{3} (x^{3} cos^{2} (x) - 3 (\frac{1}{6} x^{3} cos (2 x) - \frac{1}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{π}{2}} : [\frac{1}{24} (8 x^{3} cos^{2} (x) - 4 x^{3} cos (2 x) + 6 x^{2} sin (2 x) + 6 x cos (2 x) - 3 sin (2 x))]_{0}^{\frac{π}{2}}

= [\frac{1}{24} (8 x^{3} cos^{2} (x) - 4 x^{3} cos (2 x) + 6 x^{2} sin (2 x) + 6 x cos (2 x) - 3 sin (2 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π ^{3} - 6 π}{48}

= \frac{π ^{3} - 6 π}{48}

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\int_{1}^{3} (x^{2} - 4 x) d x

What is the integral from 0 to pi/2 of x^2cos^2(x) ?
The integral from 0 to pi/2 of x^2cos^2(x) is (pi^3-6pi)/(48)