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

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

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integral of (1-2sin(θ))^2

\int (1 - 2 sin (θ))^{2} d θ

3 θ + 4 cos (θ) - sin (2 θ) + C

Solution steps

\int (1 - 2 sin (θ))^{2} d θ

Factor

1 - 2 sin (θ) : - (- 1 + 2 sin (θ))

= \int (- (- 1 + 2 sin (θ)))^{2} d θ

Simplify

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

= \int (- 1 + 2 sin (θ))^{2} d θ

Expand

(- 1 + 2 sin (θ))^{2} : 1 - 4 sin (θ) + 4 sin^{2} (θ)

= \int 1 - 4 sin (θ) + 4 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 4 sin (θ) d θ + \int 4 sin^{2} (θ) d θ

\int 1 d θ = θ

\int 4 sin (θ) d θ = - 4 cos (θ)

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

= θ - (- 4 cos (θ)) + 2 θ - sin (2 θ)

Simplify

= 3 θ + 4 cos (θ) - sin (2 θ)

Add a constant to the solution

= 3 θ + 4 cos (θ) - sin (2 θ) + C

\int \frac{1}{( 3 + sin ( x ) )} d x

\int \frac{ln ( x )}{x ( 6 - ln ( x ) )} d x

\int y e^{x y} - \frac{1}{y} d x

\int x \cdot x^{2} + 9 d x

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