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

The integral of sqrt(cos(6x))sin(6x) is -1/9 cos^{3/2}(6x)+C

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

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

- \frac{1}{9} cos^{\frac{3}{2}} (6 x) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

= - \frac{1}{3} \cdot \int u^{2} d u

Apply the Power Rule

= - \frac{1}{3} \cdot \frac{u ^{3}}{3}

Substitute back

u = cos (6 x)

= - \frac{1}{3} \cdot \frac{( cos ( 6 x ) ) ^{3}}{3}

Simplify

- \frac{1}{3} \cdot \frac{( cos ( 6 x ) ) ^{3}}{3} : - \frac{1}{9} cos^{\frac{3}{2}} (6 x)

= - \frac{1}{9} cos^{\frac{3}{2}} (6 x)

Add a constant to the solution

= - \frac{1}{9} cos^{\frac{3}{2}} (6 x) + C

What is the integral of sqrt(cos(6x))sin(6x) ?
The integral of sqrt(cos(6x))sin(6x) is -1/9 cos^{3/2}(6x)+C