What is the integral of (1-cos(θ))^3 ?

The integral of (1-cos(θ))^3 is 5/2 θ+3/4 sin(2θ)-4sin(θ)+1/3 sin^3(θ)+C

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integral of (1-cos(θ))^3

\int (1 - cos (θ))^{3} d θ

\frac{5}{2} θ + \frac{3}{4} sin (2 θ) - 4 sin (θ) + \frac{1}{3} sin^{3} (θ) + C

Solution steps

\int (1 - cos (θ))^{3} d θ

Expand

(1 - cos (θ))^{3} : 1 - 3 cos (θ) + 3 cos^{2} (θ) - cos^{3} (θ)

= \int 1 - 3 cos (θ) + 3 cos^{2} (θ) - cos^{3} (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 3 cos (θ) d θ + \int 3 cos^{2} (θ) d θ - \int cos^{3} (θ) d θ

\int 1 d θ = θ

\int 3 cos (θ) d θ = 3 sin (θ)

\int 3 cos^{2} (θ) d θ = \frac{3}{2} (θ + \frac{1}{2} sin (2 θ))

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

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

Simplify

θ - 3 sin (θ) + \frac{3}{2} (θ + \frac{1}{2} sin (2 θ)) - (sin (θ) - \frac{sin ^{3} ( θ )}{3}) : \frac{5}{2} θ + \frac{3}{4} sin (2 θ) - 4 sin (θ) + \frac{1}{3} sin^{3} (θ)

= \frac{5}{2} θ + \frac{3}{4} sin (2 θ) - 4 sin (θ) + \frac{1}{3} sin^{3} (θ)

Add a constant to the solution

= \frac{5}{2} θ + \frac{3}{4} sin (2 θ) - 4 sin (θ) + \frac{1}{3} sin^{3} (θ) + C

x^{^{''}} + b x = 0

\int (2 x^{2} + 3 x + 7) d x

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

f (x) = 7 x^{3} e^{- x}

\int \frac{1}{y ^{\frac{1}{2}}} d y

What is the integral of (1-cos(θ))^3 ?
The integral of (1-cos(θ))^3 is 5/2 θ+3/4 sin(2θ)-4sin(θ)+1/3 sin^3(θ)+C