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

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

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

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

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{- 2 π}^{2 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)]_{- 2 π}^{2 π}

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

Simplify

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

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

Compute the boundaries:

2 π

= 2 π

\int_{1}^{2} 4 d x

\int_{0}^{l n (2)} e^{- 2 x} d x

\int_{1}^{2} (x + \frac{1}{x}) d x

\int_{0}^{1} k x^{2} d x

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

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