What is the integral of 9sin^2(x)cos^2(x) ?

The integral of 9sin^2(x)cos^2(x) is 9/8 (x-1/4 sin(4x))+C

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integral of 9sin^2(x)cos^2(x)

\int 9 sin^{2} (x) cos^{2} (x) d x

\frac{9}{8} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int 9 sin^{2} (x) cos^{2} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 9 \cdot \int \frac{1 - cos ( 4 x )}{8} d x

Take the constant out:

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

= 9 \cdot \frac{1}{8} \cdot \int 1 - cos (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{8} (\int 1 d x - \int cos (4 x) d x)

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= 9 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

9 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{9}{8} (x - \frac{1}{4} sin (4 x))

= \frac{9}{8} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{9}{8} (x - \frac{1}{4} sin (4 x)) + C

\frac{d y}{d x} = \frac{2 x y}{x ^{2} + 1}

e^{2 x} - 1

c cos (t) + t^{2} sin (t)

\int cos (2 x + 1) d x

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

What is the integral of 9sin^2(x)cos^2(x) ?
The integral of 9sin^2(x)cos^2(x) is 9/8 (x-1/4 sin(4x))+C