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

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

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

\int_{0}^{π} x^{2} cos (2 x) d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{0}^{π} x^{2} cos (2 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\int_{0}^{1} \frac{2 x}{5 x ^{2} - 3} d x

\int_{x}^{1} 2 θ sin (θ) d θ

\int_{0}^{2} \frac{6}{7} (x^{2} + \frac{x y}{2}) d y

\int_{0}^{2} x^{3} - 6 x^{2} + 8 x d x

\int_{0}^{\infty} sin (x^{2}) d x

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