What is the integral of 6xsin(x)cos(x) ?

The integral of 6xsin(x)cos(x) is 3(-1/2 xcos(2x)+1/4 sin(2x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6xsin(x)cos(x)

\int 6 x sin (x) cos (x) d x

3 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C

Solution steps

\int 6 x sin (x) cos (x) d x

Take the constant out:

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

= 6 \cdot \int x sin (x) cos (x) d x

Rewrite using trig identities

= 6 \cdot \int \frac{1}{2} x sin (2 x) d x

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int x sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} x cos (2 x) - \int - \frac{1}{2} cos (2 x) d x

= 6 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - \int - \frac{1}{2} cos (2 x) d x)

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= 6 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x)))

Simplify

6 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x))) : 3 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

= 3 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

Add a constant to the solution

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

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

y = cos^{2} (e^{c o s^{2} (t)})

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

49 y^{^{''}} + 126 y^{^{'}} + 97 y = 0

\frac{1}{d ^{7}}

What is the integral of 6xsin(x)cos(x) ?
The integral of 6xsin(x)cos(x) is 3(-1/2 xcos(2x)+1/4 sin(2x))+C