What is the integral from 0 to pi/2 of x*sin(2x) ?

The integral from 0 to pi/2 of x*sin(2x) is pi/4

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integral from 0 to pi/2 of x*sin(2x)

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int_{0}^{1} \frac{1}{x ( x + 1 )} d x

\int_{0}^{\frac{π}{4}} cos^{2} (2 x) d x

\int_{- π}^{π} x^{2} cos (x) d x

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

\int_{- 1}^{1} 2 cos (\frac{π}{3} x) d x

What is the integral from 0 to pi/2 of x*sin(2x) ?
The integral from 0 to pi/2 of x*sin(2x) is pi/4