What is the integral from pi/2 to pi of 2sin^2(θ) ?

The integral from pi/2 to pi of 2sin^2(θ) is pi-pi/2

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integral from pi/2 to pi of 2sin^2(θ)

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

π - \frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{\frac{π}{2}}^{π} sin^{2} (θ) d θ

Rewrite using trig identities

= 2 \cdot \int_{\frac{π}{2}}^{π} \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{\frac{π}{2}}^{π} 1 - cos (2 θ) d θ

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{2} (\int_{\frac{π}{2}}^{π} 1 d θ - \int_{\frac{π}{2}}^{π} cos (2 θ) d θ)

\int_{\frac{π}{2}}^{π} 1 d θ = π - \frac{π}{2}

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

= 2 \cdot \frac{1}{2} (π - \frac{π}{2} - 0)

Simplify

2 \cdot \frac{1}{2} (π - \frac{π}{2} - 0) : π - \frac{π}{2}

= π - \frac{π}{2}

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\int_{\frac{π}{2}}^{π} cos (2 θ) d θ

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\int_{0}^{kπ} x sin (x) d x

What is the integral from pi/2 to pi of 2sin^2(θ) ?
The integral from pi/2 to pi of 2sin^2(θ) is pi-pi/2