What is the integral of 9xsin(x)cos(x) ?

The integral of 9xsin(x)cos(x) is 9/2 (-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 9xsin(x)cos(x)

\int 9 x sin (x) cos (x) d x

\frac{9}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C

Solution steps

\int 9 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

= 9 \cdot \int x sin (x) cos (x) d x

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x)))

Simplify

9 \cdot \frac{1}{2} (- \frac{1}{2} x cos (2 x) - (- \frac{1}{4} sin (2 x))) : \frac{9}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

= \frac{9}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

Add a constant to the solution

= \frac{9}{2} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C

f (x) = ∣ x ∣ 5 - x^{2}, (2, 2)

x \to 2 lim (3 x + 4 x)

\int_{2}^{4} \frac{2}{x x ^{2} - 4} d x

x \to - 1 + lim (\frac{2}{x + 1})

f (x) = 5 cos (x), at x = \frac{π}{6}

What is the integral of 9xsin(x)cos(x) ?
The integral of 9xsin(x)cos(x) is 9/2 (-1/2 xcos(2x)+1/4 sin(2x))+C