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

The integral from 0 to pi/2 of sin(t)cos(t) is 1/2

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Apply u-substitution

: \int_{0}^{1} u d u

= \int_{0}^{1} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{1}

= [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int_{0}^{3} - 2 x^{2} + 6 x d x

\int_{0}^{x^{3}} sin (t^{2}) d t

\int_{0}^{4} 2^{\frac{x}{2}} d x

\int_{0}^{5} \frac{x}{4} d x

\int_{- 2}^{2} (x^{3} - 4 x) d x

What is the integral from 0 to pi/2 of sin(t)cos(t) ?
The integral from 0 to pi/2 of sin(t)cos(t) is 1/2