What is the integral of 6sin^2(3x) ?

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

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integral of 6sin^2(3x)

\int 6 sin^{2} (3 x) d x

3 (x - \frac{1}{6} sin (6 x)) + C

Solution steps

\int 6 sin^{2} (3 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} (3 x) d x

Rewrite using trig identities

= 6 \cdot \int \frac{1}{2} (1 - cos (6 x)) d x

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int 1 - cos (6 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}{2} (\int 1 d x - \int cos (6 x) d x)

\int 1 d x = x

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

= 6 \cdot \frac{1}{2} (x - \frac{1}{6} sin (6 x))

Simplify

6 \cdot \frac{1}{2} (x - \frac{1}{6} sin (6 x)) : 3 (x - \frac{1}{6} sin (6 x))

= 3 (x - \frac{1}{6} sin (6 x))

Add a constant to the solution

= 3 (x - \frac{1}{6} sin (6 x)) + C

f (x) = \frac{1 - 5 x}{1 + 7 x}

\frac{d}{d x} (cot (3 x))

\int sin^{7} (x) cos (x) d x

n = 0 \sum \infty n^{2} e^{- n}

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

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