What is the integral from pi/2 to 0 of-sin^2(t) ?

The integral from pi/2 to 0 of-sin^2(t) is pi/4

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integral from pi/2 to 0 of-sin^2(t)

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{\frac{π}{2}} - sin^{2} (t) d t

Take the constant out:

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

= - (- \int_{0}^{\frac{π}{2}} sin^{2} (t) d t)

Rewrite using trig identities

= - (- \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 t )}{2} d t)

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (2 t) d t)

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_{0}^{\frac{π}{2}} 1 d t - \int_{0}^{\frac{π}{2}} cos (2 t) d t))

\int_{0}^{\frac{π}{2}} 1 d t = \frac{π}{2}

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

= - (- \frac{1}{2} (\frac{π}{2} - 0))

Simplify

- (- \frac{1}{2} (\frac{π}{2} - 0)) : \frac{π}{4}

= \frac{π}{4}

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What is the integral from pi/2 to 0 of-sin^2(t) ?
The integral from pi/2 to 0 of-sin^2(t) is pi/4