What is the integral from 0 to pi/2 of (x+1)cos(2x) ?

The integral from 0 to pi/2 of (x+1)cos(2x) is -1/2

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integral from 0 to pi/2 of (x+1)cos(2x)

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{1}{2}

= - \frac{1}{2}

\int_{0}^{2 π} 1 + cos (x) d x

\int_{0}^{1} 3^{x} e^{x} sin (x) d x

\int_{3}^{5} - x^{2} + 7 x - 10 d x

\int_{0}^{2 π} cos^{2} (θ) sin (θ) d θ

\int_{- π}^{π} e^{x} cos (x) d x

What is the integral from 0 to pi/2 of (x+1)cos(2x) ?
The integral from 0 to pi/2 of (x+1)cos(2x) is -1/2