What is the integral of 19sin(x)sin(2x) ?

The integral of 19sin(x)sin(2x) is 38/3 sin^3(x)+C

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integral of 19sin(x)sin(2x)

\int 19 sin (x) sin (2 x) d x

\frac{38}{3} sin^{3} (x) + C

Solution steps

\int 19 sin (x) sin (2 x) d x

Take the constant out:

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

= 19 \cdot \int sin (x) sin (2 x) d x

Rewrite using trig identities

= 19 \cdot \int 2 sin^{2} (x) cos (x) d x

Take the constant out:

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

= 19 \cdot 2 \cdot \int sin^{2} (x) cos (x) d x

Apply u-substitution

: \int u^{2} d u

= 19 \cdot 2 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 19 \cdot 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = sin (x)

= 19 \cdot 2 \cdot \frac{sin ^{3} ( x )}{3}

Simplify

19 \cdot 2 \cdot \frac{sin ^{3} ( x )}{3} : \frac{38}{3} sin^{3} (x)

= \frac{38}{3} sin^{3} (x)

Add a constant to the solution

= \frac{38}{3} sin^{3} (x) + C

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What is the integral of 19sin(x)sin(2x) ?
The integral of 19sin(x)sin(2x) is 38/3 sin^3(x)+C