What is the integral from 0 to t of sin(u)sin(2u) ?

The integral from 0 to t of sin(u)sin(2u) is (2sin^3(t))/3

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integral from 0 to t of sin(u)sin(2u)

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

\frac{2 sin ^{3} ( t )}{3}

Solution steps

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

Rewrite using trig identities

= \int_{0}^{t} 2 sin^{2} (u) cos (u) d u

Take the constant out:

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

= 2 \cdot \int_{0}^{t} sin^{2} (u) cos (u) d u

Apply u-substitution

: \int_{0}^{s i n (t)} v^{2} d v

= 2 \cdot \int_{0}^{s i n (t)} v^{2} d v

Apply the Power Rule

: [\frac{v ^{3}}{3}]_{0}^{s i n (t)}

= 2 [\frac{v ^{3}}{3}]_{0}^{s i n (t)}

Compute the boundaries:

\frac{sin ^{3} ( t )}{3}

= 2 \cdot \frac{sin ^{3} ( t )}{3}

Simplify

= \frac{2 sin ^{3} ( t )}{3}

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What is the integral from 0 to t of sin(u)sin(2u) ?
The integral from 0 to t of sin(u)sin(2u) is (2sin^3(t))/3