What is the integral of 9cos(x)sin(x) ?

The integral of 9cos(x)sin(x) is 9/2 sin^2(x)+C

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integral of 9cos(x)sin(x)

\int 9 cos (x) sin (x) d x

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

Solution steps

\int 9 cos (x) sin (x) d x

Take the constant out:

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

= 9 \cdot \int cos (x) sin (x) d x

Apply u-substitution

: \int u d u

= 9 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 9 \cdot \frac{u ^{2}}{2}

Substitute back

u = sin (x)

= 9 \cdot \frac{sin ^{2} ( x )}{2}

Simplify

9 \cdot \frac{sin ^{2} ( x )}{2} : \frac{9}{2} sin^{2} (x)

= \frac{9}{2} sin^{2} (x)

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2}}{7})

f (x) = 2 x^{3} - 4 x^{2}, at x = 1

\int_{1}^{5} \frac{5}{x ^{2} + 4 x + 4} d x

\frac{d}{d x} (x^{2} + 2 x - 3)

\frac{\partial}{\partial y} (\frac{x ^{2}}{1 + y ^{3}})

What is the integral of 9cos(x)sin(x) ?
The integral of 9cos(x)sin(x) is 9/2 sin^2(x)+C