What is the integral of (2-sin(θ))^2 ?

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

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integral of (2-sin(θ))^2

\int (2 - sin (θ))^{2} d θ

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

Solution steps

\int (2 - sin (θ))^{2} d θ

Expand

(2 - sin (θ))^{2} : 4 - 4 sin (θ) + sin^{2} (θ)

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

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 θ - \int 4 sin (θ) d θ + \int sin^{2} (θ) d θ

\int 4 d θ = 4 θ

\int 4 sin (θ) d θ = - 4 cos (θ)

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

= 4 θ - (- 4 cos (θ)) + \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

4 θ - (- 4 cos (θ)) + \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : \frac{9}{2} θ - \frac{1}{4} sin (2 θ) + 4 cos (θ)

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

Add a constant to the solution

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

\frac{2}{( s + 1 ) ^{2}}

\int 6 cos^{2} (x) d x

y^{^{''}} + 7 y^{^{'}} - 18 y = 0

\frac{x - 5 x x}{x}

x \to 5 lim (\frac{12}{5 - x})

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