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

The integral of (sqrt(cos(θ))-2sin(θ))^2 is sin(θ)+8/3 cos^{3/2}(θ)+2θ-sin(2θ)+C

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integral of (sqrt(cos(θ))-2sin(θ))^2

\int (cos (θ) - 2 sin (θ))^{2} d θ

sin (θ) + \frac{8}{3} cos^{\frac{3}{2}} (θ) + 2 θ - sin (2 θ) + C

Solution steps

\int (cos (θ) - 2 sin (θ))^{2} d θ

Expand

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

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

\int cos (θ) d θ = sin (θ)

\int 4 sin (θ) cos (θ) d θ = - \frac{8}{3} cos^{\frac{3}{2}} (θ)

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

= sin (θ) - (- \frac{8}{3} cos^{\frac{3}{2}} (θ)) + 2 θ - sin (2 θ)

Simplify

= sin (θ) + \frac{8}{3} cos^{\frac{3}{2}} (θ) + 2 θ - sin (2 θ)

Add a constant to the solution

= sin (θ) + \frac{8}{3} cos^{\frac{3}{2}} (θ) + 2 θ - sin (2 θ) + C

(x - y + 2) d x - (x - y) d y = 0

x \to 0 lim ((1 + 2 x)^{\frac{5}{x}})

2 x^{3} + 2 x y + 2 y^{2} = 38, (- 1, - 4)

x \to 0 lim (\frac{sin ( x )}{x} - 1)

\int_{2}^{5} \frac{1}{x - 2} d x

What is the integral of (sqrt(cos(θ))-2sin(θ))^2 ?
The integral of (sqrt(cos(θ))-2sin(θ))^2 is sin(θ)+8/3 cos^{3/2}(θ)+2θ-sin(2θ)+C