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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{9} sin^{3} (3 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int v^{2} d v

= 2 \cdot \frac{1}{3} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= 2 \cdot \frac{1}{3} \cdot \frac{v ^{3}}{3}

Substitute back

= 2 \cdot \frac{1}{3} \cdot \frac{sin ^{3} ( 3 x )}{3}

Simplify

2 \cdot \frac{1}{3} \cdot \frac{sin ^{3} ( 3 x )}{3} : \frac{2}{9} sin^{3} (3 x)

= \frac{2}{9} sin^{3} (3 x)

Add a constant to the solution

= \frac{2}{9} sin^{3} (3 x) + C

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

\int csc^{6} (\frac{2 x}{3}) d x

\int (2 + 5 x) e^{\frac{1}{3} x} d x

\int \frac{36}{x ^{2} + 9} d x

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