What is the integral of (1-sin(θ))^2 ?

The integral of (1-sin(θ))^2 is 3/2 θ-1/4 sin(2θ)+2cos(θ)+C

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integral of (1-sin(θ))^2

\int (1 - sin (θ))^{2} d θ

\frac{3}{2} θ - \frac{1}{4} sin (2 θ) + 2 cos (θ) + C

Solution steps

\int (1 - sin (θ))^{2} d θ

Expand

(1 - sin (θ))^{2} : 1 - 2 sin (θ) + sin^{2} (θ)

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

\int 1 d θ = θ

\int 2 sin (θ) d θ = - 2 cos (θ)

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

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

Simplify

θ - (- 2 cos (θ)) + \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : \frac{3}{2} θ - \frac{1}{4} sin (2 θ) + 2 cos (θ)

= \frac{3}{2} θ - \frac{1}{4} sin (2 θ) + 2 cos (θ)

Add a constant to the solution

= \frac{3}{2} θ - \frac{1}{4} sin (2 θ) + 2 cos (θ) + C

\int cos (- 1 (x)) d x

\int 2 x (x^{2} + 4)^{5} d x

y = \frac{6 x}{e ^{x}}

\frac{d}{d x} ((x - 1)^{\frac{3}{2}})

\frac{d}{d x} (\frac{x ^{3} + 2}{cos ( x )})

What is the integral of (1-sin(θ))^2 ?
The integral of (1-sin(θ))^2 is 3/2 θ-1/4 sin(2θ)+2cos(θ)+C