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

x \to - 2 + lim (x^{2} - 5)

y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\int_{0}^{π} e^{- x} cos (2 x) d x

\frac{\partial}{\partial y} (ln (x + e^{- y}))

\int cos (a ln (\frac{x}{b})) d 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