What is the integral of (1+cos(θ))^2 ?

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

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integral of (1+cos(θ))^2

\int (1 + cos (θ))^{2} d θ

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

Solution steps

\int (1 + cos (θ))^{2} d θ

Expand

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

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

\int 1 d θ = θ

\int 2 cos (θ) d θ = 2 sin (θ)

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

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{1}{x}, y = x, y = \frac{1}{4} x

\int_{1}^{6} \frac{x ^{2} + 5}{7 x - x ^{2}} d x

\frac{cos ( 7 t )}{e ^{5 t}}

x \frac{d y}{d x} + y = y^{2} x^{2} ln (x)

\int \frac{y}{x ^{2} + y ^{2}} d x

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