What is the integral of (5-sin(3x))^2 ?

The integral of (5-sin(3x))^2 is 51/2 x+10/3 cos(3x)-1/12 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 (5-sin(3x))^2

\int (5 - sin (3 x))^{2} d x

\frac{51}{2} x + \frac{10}{3} cos (3 x) - \frac{1}{12} sin (6 x) + C

Solution steps

\int (5 - sin (3 x))^{2} d x

Expand

(5 - sin (3 x))^{2} : 25 - 10 sin (3 x) + sin^{2} (3 x)

= \int 25 - 10 sin (3 x) + sin^{2} (3 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

= \int 25 d x - \int 10 sin (3 x) d x + \int sin^{2} (3 x) d x

\int 25 d x = 25 x

\int 10 sin (3 x) d x = - \frac{10}{3} cos (3 x)

\int sin^{2} (3 x) d x = \frac{1}{2} (x - \frac{1}{6} sin (6 x))

= 25 x - (- \frac{10}{3} cos (3 x)) + \frac{1}{2} (x - \frac{1}{6} sin (6 x))

Simplify

25 x - (- \frac{10}{3} cos (3 x)) + \frac{1}{2} (x - \frac{1}{6} sin (6 x)) : \frac{51}{2} x + \frac{10}{3} cos (3 x) - \frac{1}{12} sin (6 x)

= \frac{51}{2} x + \frac{10}{3} cos (3 x) - \frac{1}{12} sin (6 x)

Add a constant to the solution

= \frac{51}{2} x + \frac{10}{3} cos (3 x) - \frac{1}{12} sin (6 x) + C

\int z sin (yz) d y

\int \frac{a + 4}{4 a + 7} d a

\int cot (3 s) csc^{4} (3 s) d s

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

\int (x + 4)^{6} d x

What is the integral of (5-sin(3x))^2 ?
The integral of (5-sin(3x))^2 is 51/2 x+10/3 cos(3x)-1/12 sin(6x)+C