What is the integral of 6sqrt(cos(3x))sin(3x) ?

The integral of 6sqrt(cos(3x))sin(3x) is -4/3 cos^{3/2}(3x)+C

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integral of 6sqrt(cos(3x))sin(3x)

\int 6 cos (3 x) sin (3 x) d x

- \frac{4}{3} cos^{\frac{3}{2}} (3 x) + C

Solution steps

\int 6 cos (3 x) sin (3 x) d x

Take the constant out:

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

= 6 \cdot \int cos (3 x) sin (3 x) d x

Apply u-substitution

: \int - \frac{2 u ^{2}}{3} d u

= 6 \cdot \int - \frac{2 u ^{2}}{3} d u

Take the constant out:

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

= 6 (- \frac{2}{3} \cdot \int u^{2} d u)

Apply the Power Rule

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

= 6 (- \frac{2}{3} \cdot \frac{u ^{3}}{3})

Substitute back

u = cos (3 x)

= 6 - \frac{2}{3} \cdot \frac{( cos ( 3 x ) ) ^{3}}{3}

Simplify

6 - \frac{2}{3} \cdot \frac{( cos ( 3 x ) ) ^{3}}{3} : - \frac{4}{3} cos^{\frac{3}{2}} (3 x)

= - \frac{4}{3} cos^{\frac{3}{2}} (3 x)

Add a constant to the solution

= - \frac{4}{3} cos^{\frac{3}{2}} (3 x) + C

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What is the integral of 6sqrt(cos(3x))sin(3x) ?
The integral of 6sqrt(cos(3x))sin(3x) is -4/3 cos^{3/2}(3x)+C