What is the integral of xsin^2(9x) ?

The integral of xsin^2(9x) is 1/648 (162x^2-18xsin(18x)-cos(18x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xsin^2(9x)

\int x sin^{2} (9 x) d x

\frac{1}{648} (162 x^{2} - 18 x sin (18 x) - cos (18 x)) + C

Solution steps

\int x sin^{2} (9 x) d x

Rewrite using trig identities

= \int \frac{1}{2} x (1 - cos (18 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 (18 x)) d x

Apply Integration By Parts

: x (x - \frac{1}{18} sin (18 x)) - \int x - \frac{1}{18} sin (18 x) d x

= \frac{1}{2} (x (x - \frac{1}{18} sin (18 x)) - \int x - \frac{1}{18} sin (18 x) d x)

\int x - \frac{1}{18} sin (18 x) d x = \frac{x ^{2}}{2} + \frac{1}{324} cos (18 x)

= \frac{1}{2} (x (x - \frac{1}{18} sin (18 x)) - (\frac{x ^{2}}{2} + \frac{1}{324} cos (18 x)))

Simplify

\frac{1}{2} (x (x - \frac{1}{18} sin (18 x)) - (\frac{x ^{2}}{2} + \frac{1}{324} cos (18 x))) : \frac{1}{648} (162 x^{2} - 18 x sin (18 x) - cos (18 x))

= \frac{1}{648} (162 x^{2} - 18 x sin (18 x) - cos (18 x))

Add a constant to the solution

= \frac{1}{648} (162 x^{2} - 18 x sin (18 x) - cos (18 x)) + C

2 x^{\frac{3}{2}}

\int sin^{3} (5 x) cos^{- 2} (5 x) d x

\int_{1}^{2} \frac{x ^{2} + 7}{3 x - x ^{2}} d x

\frac{s}{( 2 s + 1 ) ( ( s + 2 ) ^{2} - 1 )}

f (x) = x^{3} - x, (- 1, 0)

What is the integral of xsin^2(9x) ?
The integral of xsin^2(9x) is 1/648 (162x^2-18xsin(18x)-cos(18x))+C