What is the integral of 2sin^2(θ) ?

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

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integral of 2sin^2(θ)

\int 2 sin^{2} (θ) d θ

θ - \frac{1}{2} sin (2 θ) + C

Solution steps

\int 2 sin^{2} (θ) d θ

Take the constant out:

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

= 2 \cdot \int sin^{2} (θ) d θ

Rewrite using trig identities

= 2 \cdot \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

= 2 \cdot \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

= 2 \cdot \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)

= 2 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

2 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) : θ - \frac{1}{2} sin (2 θ)

= θ - \frac{1}{2} sin (2 θ)

Add a constant to the solution

= θ - \frac{1}{2} sin (2 θ) + C

\int_{- 6}^{6} \frac{1}{2} (36 - x^{2}) d x

\int_{- π}^{π} sin (2 x) sin (3 x) d x

f (x) = e^{x^{5}}

\int \frac{10}{10 + e ^{x}} d x

\frac{d}{d x} (\frac{( x + y )}{( x - y )})

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