What is the integral of (cos(3x))^4 ?

The integral of (cos(3x))^4 is 1/3 cos^3(3x)sin(3x)+1/32 (12x-sin(12x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(3x))^4

\int (cos (3 x))^{4} d x

\frac{1}{3} cos^{3} (3 x) sin (3 x) + \frac{1}{32} (12 x - sin (12 x)) + C

Solution steps

\int (cos (3 x))^{4} d x

Simplify

= \int cos^{4} (3 x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{3} d u

= \int cos^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{3} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= \frac{1}{3} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{3} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 3 x

= \frac{1}{3} (cos^{3} (3 x) sin (3 x) - (- \frac{3}{8} (3 x - \frac{1}{4} sin (4 \cdot 3 x))))

Simplify

\frac{1}{3} (cos^{3} (3 x) sin (3 x) - (- \frac{3}{8} (3 x - \frac{1}{4} sin (4 \cdot 3 x)))) : \frac{1}{3} cos^{3} (3 x) sin (3 x) + \frac{1}{32} (12 x - sin (12 x))

= \frac{1}{3} cos^{3} (3 x) sin (3 x) + \frac{1}{32} (12 x - sin (12 x))

Add a constant to the solution

= \frac{1}{3} cos^{3} (3 x) sin (3 x) + \frac{1}{32} (12 x - sin (12 x)) + C

\int cos (9 x + 2) d x

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

\int \frac{csc ( ln ( x ) ) cot ( ln ( x ) )}{x} d x

\int \frac{1}{1 + a ^{2} y ^{2}} d y

\int 9 sin (6 x) d x

What is the integral of (cos(3x))^4 ?
The integral of (cos(3x))^4 is 1/3 cos^3(3x)sin(3x)+1/32 (12x-sin(12x))+C