What is the integral of 18sin^2(x)cos^2(x) ?

The integral of 18sin^2(x)cos^2(x) is 9/4 (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 18sin^2(x)cos^2(x)

\int 18 sin^{2} (x) cos^{2} (x) d x

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

Solution steps

\int 18 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

= 18 \cdot \int sin^{2} (x) cos^{2} (x) d x

Rewrite using trig identities

= 18 \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

= 18 \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

= 18 \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)

= 18 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

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

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

Add a constant to the solution

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

\frac{( 2 s + 2 )}{( s ^{2} + 13 s + 12 )}

\int - \frac{4 x}{( 1 - 2 x ^{2} ) ^{3}} d x

\int (x^{\frac{1}{3}} - x^{\frac{2}{3}}) d x

\int (x^{7}) d x

\frac{\partial}{\partial y} (4 - x^{2} - 3 y^{2})

What is the integral of 18sin^2(x)cos^2(x) ?
The integral of 18sin^2(x)cos^2(x) is 9/4 (x-1/4 sin(4x))+C