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

The integral of (cos(x)+2sin(x))^2 is 5/2 x-3/4 sin(2x)+2sin^2(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)+2sin(x))^2

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

\frac{5}{2} x - \frac{3}{4} sin (2 x) + 2 sin^{2} (x) + C

Solution steps

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

Expand

(cos (x) + 2 sin (x))^{2} : cos^{2} (x) + 4 cos (x) sin (x) + 4 sin^{2} (x)

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

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

\int 4 cos (x) sin (x) d x = 2 sin^{2} (x)

\int 4 sin^{2} (x) d x = 2 x - sin (2 x)

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

Simplify

\frac{1}{2} (x + \frac{1}{2} sin (2 x)) + 2 sin^{2} (x) + 2 x - sin (2 x) : \frac{5}{2} x - \frac{3}{4} sin (2 x) + 2 sin^{2} (x)

= \frac{5}{2} x - \frac{3}{4} sin (2 x) + 2 sin^{2} (x)

Add a constant to the solution

= \frac{5}{2} x - \frac{3}{4} sin (2 x) + 2 sin^{2} (x) + C

\int \frac{x ^{2} + 3 x + 8}{x ^{3} - x ^{2} + 3 x - 3} d x

\int \frac{z ^{4} - 3 z ^{2} - 4 z}{5 z} d z

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

\int x + \frac{3 y}{x} d x

\int 16 cos^{4} (5 π) t d t

What is the integral of (cos(x)+2sin(x))^2 ?
The integral of (cos(x)+2sin(x))^2 is 5/2 x-3/4 sin(2x)+2sin^2(x)+C