What is the integral of 9cos^2(3x) ?

The integral of 9cos^2(3x) is 9/2 (x+1/6 sin(6x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9cos^2(3x)

\int 9 cos^{2} (3 x) d x

\frac{9}{2} (x + \frac{1}{6} sin (6 x)) + C

Solution steps

\int 9 cos^{2} (3 x) d x

Take the constant out:

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

= 9 \cdot \int cos^{2} (3 x) d x

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (x + \frac{1}{6} sin (6 x))

Simplify

9 \cdot \frac{1}{2} (x + \frac{1}{6} sin (6 x)) : \frac{9}{2} (x + \frac{1}{6} sin (6 x))

= \frac{9}{2} (x + \frac{1}{6} sin (6 x))

Add a constant to the solution

= \frac{9}{2} (x + \frac{1}{6} sin (6 x)) + C

\int \frac{1}{5 + cos ( x )} d x

\int \frac{x ^{3}}{x ^{2} - 3 x + 2} d x

\int (x^{5} + \frac{1}{x ^{3}} + 4) d x

\int - 3 x^{4} e^{x^{5}} d x

\int x^{\frac{- 1}{3}} d x

What is the integral of 9cos^2(3x) ?
The integral of 9cos^2(3x) is 9/2 (x+1/6 sin(6x))+C