What is the integral of sin^2(φ) ?

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

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integral of sin^2(φ)

\int sin^{2} (φ) d φ

\frac{1}{2} (φ - \frac{1}{2} sin (2 φ)) + C

Solution steps

\int sin^{2} (φ) d φ

Rewrite using trig identities

= \int \frac{1 - cos ( 2 φ )}{2} d φ

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 φ) d φ

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 φ - \int cos (2 φ) d φ)

\int 1 d φ = φ

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

= \frac{1}{2} (φ - \frac{1}{2} sin (2 φ))

Add a constant to the solution

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

f (x) = 4 x^{(2)} (x - 3)

x \to \infty lim (x^{2} + 2 x)

\frac{d}{d x} (arctan (\frac{8}{x}))

\int \frac{4}{( x + 3 ) ^{2}} d x

\frac{\partial}{\partial y} (x^{3} y^{10} sin (x y^{2}))

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