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

The integral from pi/4 to pi/2 of 2x-sin(2x) is (3pi^2-8)/(16)

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integral from pi/4 to pi/2 of 2x-sin(2x)

\int_{\frac{π}{4}}^{\frac{π}{2}} 2 x - sin (2 x) d x

\frac{3 π ^{2} - 8}{16}

Decimal Notation

1.35055 \dots

Solution steps

\int_{\frac{π}{4}}^{\frac{π}{2}} 2 x - sin (2 x) d x

Apply u-substitution

: \int_{\frac{π}{2}}^{π} (u - sin (u)) \frac{1}{2} d u

= \int_{\frac{π}{2}}^{π} (u - sin (u)) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{\frac{π}{2}}^{π} u - sin (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{\frac{π}{2}}^{π} u d u - \int_{\frac{π}{2}}^{π} sin (u) d u)

\int_{\frac{π}{2}}^{π} u d u = \frac{3 π ^{2}}{8}

\int_{\frac{π}{2}}^{π} sin (u) d u = 1

= \frac{1}{2} (\frac{3 π ^{2}}{8} - 1)

Simplify

\frac{1}{2} (\frac{3 π ^{2}}{8} - 1) : \frac{3 π ^{2} - 8}{16}

= \frac{3 π ^{2} - 8}{16}

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What is the integral from pi/4 to pi/2 of 2x-sin(2x) ?
The integral from pi/4 to pi/2 of 2x-sin(2x) is (3pi^2-8)/(16)