What is the integral of 6sin^2(x)cos^2(x) ?

The integral of 6sin^2(x)cos^2(x) is 3/4 (x-1/4 sin(4x))+C

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integral of 6sin^2(x)cos^2(x)

\int 6 sin^{2} (x) cos^{2} (x) d x

\frac{3}{4} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int 6 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

= 6 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 6 \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

= 6 \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

= 6 \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)

= 6 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

6 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{3}{4} (x - \frac{1}{4} sin (4 x))

= \frac{3}{4} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{3}{4} (x - \frac{1}{4} sin (4 x)) + C

n = 1 \sum \infty \frac{3}{2 n}

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

(ln (x^{2} + 8 x + 21))^{^{'}}

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

\int \frac{( x ^{3} - 2 x ^{2} - 4 )}{( x ^{3} - 2 x ^{2} )} d x

What is the integral of 6sin^2(x)cos^2(x) ?
The integral of 6sin^2(x)cos^2(x) is 3/4 (x-1/4 sin(4x))+C