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

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

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integral of (2+cos(x))^2

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

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

Solution steps

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

Expand

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

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

\int 4 d x = 4 x

\int 4 cos (x) d x = 4 sin (x)

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

= 4 x + 4 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

4 x + 4 sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{9}{2} x + \frac{1}{4} sin (2 x) + 4 sin (x)

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

Add a constant to the solution

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

\frac{( - x ^{2} + 2 x + 2 )}{( x + 1 ) ^{4}}

y^{^{''}} + y^{^{'}} - 6 y = 2 e^{5 x}

\frac{\partial}{\partial x} (y e^{(x^{2} - y)})

y = 29 x

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

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