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

The integral of 9xsin^2(x) is 9(1/2 x(x-1/2 sin(2x))-1/2 ((x^2)/2+1/4 cos(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^2(x)

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

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

Solution steps

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

Apply Integration By Parts

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

= 9 (\frac{1}{2} x (x - \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)

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

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

Add a constant to the solution

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

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

x \to 2 lim ([- \frac{x}{2}] + x^{2})

\frac{\partial}{\partial y} (3 sin (x) sin (y))

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

e^{x}, 3

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