What is the integral of xcos^2(6x) ?

The integral of xcos^2(6x) is 1/288 (72x^2+12xsin(12x)+cos(12x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xcos^2(6x)

\int x cos^{2} (6 x) d x

\frac{1}{288} (72 x^{2} + 12 x sin (12 x) + cos (12 x)) + C

Solution steps

\int x cos^{2} (6 x) d x

Rewrite using trig identities

= \int \frac{1}{2} x (1 + cos (12 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x (1 + cos (12 x)) d x

Apply Integration By Parts

: x (x + \frac{1}{12} sin (12 x)) - \int x + \frac{1}{12} sin (12 x) d x

= \frac{1}{2} (x (x + \frac{1}{12} sin (12 x)) - \int x + \frac{1}{12} sin (12 x) d x)

\int x + \frac{1}{12} sin (12 x) d x = \frac{x ^{2}}{2} - \frac{1}{144} cos (12 x)

= \frac{1}{2} (x (x + \frac{1}{12} sin (12 x)) - (\frac{x ^{2}}{2} - \frac{1}{144} cos (12 x)))

Simplify

\frac{1}{2} (x (x + \frac{1}{12} sin (12 x)) - (\frac{x ^{2}}{2} - \frac{1}{144} cos (12 x))) : \frac{1}{288} (72 x^{2} + 12 x sin (12 x) + cos (12 x))

= \frac{1}{288} (72 x^{2} + 12 x sin (12 x) + cos (12 x))

Add a constant to the solution

= \frac{1}{288} (72 x^{2} + 12 x sin (12 x) + cos (12 x)) + C

\frac{d}{d x} (0.5 x^{2} e^{- 0.6 x})

x ln (x) y^{^{'}} + y = 8 x e^{x}

\frac{x}{1 - x ^{2}}

n = 0 \sum \infty \frac{1}{n ln ( n )}

\frac{\partial}{\partial x} (x^{2} y + 2 y^{2} - 4 x y + 6)

What is the integral of xcos^2(6x) ?
The integral of xcos^2(6x) is 1/288 (72x^2+12xsin(12x)+cos(12x))+C