What is the integral of 1-sin^2(θ) ?

The integral of 1-sin^2(θ) is 1/2 (θ+1/2 sin(2θ))+C

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integral of 1-sin^2(θ)

\int 1 - sin^{2} (θ) d θ

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

Solution steps

\int 1 - sin^{2} (θ) d θ

Rewrite using trig identities

= \int \frac{1 + cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 + 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

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

\int 1 d θ = θ

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

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

Add a constant to the solution

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

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\int 4 y sin (x y) d x

\frac{d}{d x} (A (- 2 e^{- 2 x} x + e^{- 2 x}))

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\frac{\partial}{\partial x} (x^{3} - 2 x y)

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