What is the integral of 19sin^2(x)cos^2(x) ?

The integral of 19sin^2(x)cos^2(x) is 19/8 (x-1/4 sin(4x))+C

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integral of 19sin^2(x)cos^2(x)

\int 19 sin^{2} (x) cos^{2} (x) d x

\frac{19}{8} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int 19 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

= 19 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 19 \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

= 19 \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

= 19 \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)

= 19 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

19 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{19}{8} (x - \frac{1}{4} sin (4 x))

= \frac{19}{8} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{19}{8} (x - \frac{1}{4} sin (4 x)) + C

y = \frac{x + 1}{2 x - 1}

y^{^{'}} - y = e^{2 x}

\int 12 cos (4 x) d x

f (x) = 5 x + 1

x \to 0 - lim (\frac{x ^{2}}{6} - \frac{5}{x})

What is the integral of 19sin^2(x)cos^2(x) ?
The integral of 19sin^2(x)cos^2(x) is 19/8 (x-1/4 sin(4x))+C