What is the integral of-cos^{1/2}(3x)sin(3x) ?

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

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integral of-cos^{1/2}(3x)sin(3x)

\int - cos^{\frac{1}{2}} (3 x) sin (3 x) d x

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

Solution steps

\int - 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

= - \int cos (3 x) sin (3 x) d x

Apply u-substitution

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

= - \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

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

Apply the Power Rule

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

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

Substitute back

u = cos (3 x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (y x^{y - 1})

y^{^{''}} + 6 y^{^{'}} + 58 y = 0

\frac{d}{d x} (\frac{- x ^{2}}{8})

\int \frac{1}{( c ^{2} + x ^{2} )} d x

\frac{d}{d t} (2 e^{t})

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