What is the integral of 12sin^2(x)cos^2(x) ?

The integral of 12sin^2(x)cos^2(x) is 3/2 (x-1/4 sin(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 12sin^2(x)cos^2(x)

\int 12 sin^{2} (x) cos^{2} (x) d x

\frac{3}{2} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int 12 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

= 12 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 12 \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

= 12 \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

= 12 \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)

= 12 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

12 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{3}{2} (x - \frac{1}{4} sin (4 x))

= \frac{3}{2} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{3}{2} (x - \frac{1}{4} sin (4 x)) + C

cos (x) x^{3}

\int_{- π}^{π} cos (n x) cos (m x) d x

y^{^{''}} - 12 y^{^{'}} + 36 y = t e^{6 t}

\int sin (\frac{nπ x}{4}) d x

\frac{d}{d x} (\frac{2 + sin ( x )}{2 x + cos ( x )})

What is the integral of 12sin^2(x)cos^2(x) ?
The integral of 12sin^2(x)cos^2(x) is 3/2 (x-1/4 sin(4x))+C