What is the integral of+(8x+3sin(x))^2 ?

The integral of+(8x+3sin(x))^2 is (64x^3)/3+48(-xcos(x)+sin(x))+9/2 (x-1/2 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+(8x+3sin(x))^2

\int + (8 x + 3 sin (x))^{2} d x

\frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

\int (8 x + 3 sin (x))^{2} d x

Expand

(8 x + 3 sin (x))^{2} : 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x)

= \int 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 64 x^{2} d x + \int 48 x sin (x) d x + \int 9 sin^{2} (x) d x

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

\int 48 x sin (x) d x = 48 (- x cos (x) + sin (x))

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

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

4 x^{2} + 1

4 y^{^{''}} - 4 y^{^{'}} + 10 y = 0

\int \frac{x}{( x ^{2} + 1 ) ( x - 1 )} d x

\frac{d}{d x} (x^{(x^{x})})

x \to 6 lim (\frac{x ^{2}}{x - 6})

What is the integral of+(8x+3sin(x))^2 ?
The integral of+(8x+3sin(x))^2 is (64x^3)/3+48(-xcos(x)+sin(x))+9/2 (x-1/2 sin(2x))+C