What is the integral of (4-3sin(2x))^4cos(2x) ?

The integral of (4-3sin(2x))^4cos(2x) is -1/30 (4-3sin(2x))^5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4-3sin(2x))^4cos(2x)

\int (4 - 3 sin (2 x))^{4} cos (2 x) d x

- \frac{1}{30} (4 - 3 sin (2 x))^{5} + C

Solution steps

\int (4 - 3 sin (2 x))^{4} cos (2 x) d x

Apply u-substitution

: \int - \frac{u ^{4}}{6} d u

= \int - \frac{u ^{4}}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= - \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 4 - 3 sin (2 x)

= - \frac{1}{6} \cdot \frac{( 4 - 3 sin ( 2 x ) ) ^{5}}{5}

Simplify

- \frac{1}{6} \cdot \frac{( 4 - 3 sin ( 2 x ) ) ^{5}}{5} : - \frac{1}{30} (4 - 3 sin (2 x))^{5}

= - \frac{1}{30} (4 - 3 sin (2 x))^{5}

Add a constant to the solution

= - \frac{1}{30} (4 - 3 sin (2 x))^{5} + C

h (a) = 2 x - 5

ln (\frac{s + 5}{s ^{2} + 2 s + 6})

\frac{\partial}{\partial x} (3 x^{4})

\frac{d}{d x} ((\frac{x ^{3}}{x ^{8} + 9})^{2})

2 y y^{^{'}} + 2 = y^{2} + 2 x

What is the integral of (4-3sin(2x))^4cos(2x) ?
The integral of (4-3sin(2x))^4cos(2x) is -1/30 (4-3sin(2x))^5+C