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

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

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integral from 0 to pi/2 of 2xsin(2x)

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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}}

= 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)

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Simplify

= \frac{π}{2}

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\int_{0}^{13} ∣ x - 10 ∣ d x

\int_{0}^{t a n (θ)} sec^{2} (y) d y

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