What is the integral of (cos(x)+1)^2-(sin(x)+1)^2 ?

The integral of (cos(x)+1)^2-(sin(x)+1)^2 is 1/2 sin(2x)+2sin(x)+2cos(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(x)+1)^2-(sin(x)+1)^2

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

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

Solution steps

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

Expand

(cos (x) + 1)^{2} - (sin (x) + 1)^{2} : cos^{2} (x) + 2 cos (x) - sin^{2} (x) - 2 sin (x)

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

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

\int 2 cos (x) d x = 2 sin (x)

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

\int 2 sin (x) d x = - 2 cos (x)

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

Simplify

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

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

Add a constant to the solution

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

- 2 (\frac{d y}{d x}) + 6 x^{2} = 0

y^{^{''}} + 25 y = - 40 sec (5 t)

\int \frac{1}{a ^{2} - x ^{2}} d x

x \to 2 lim ((x^{2})^{2})

y = x, y = 3 x, y = - x + 2

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