What is the integral of sin^2(5θ) ?

The integral of sin^2(5θ) is 1/2 (θ-1/10 sin(10θ))+C

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integral of sin^2(5θ)

\int sin^{2} (5 θ) d θ

\frac{1}{2} (θ - \frac{1}{10} sin (10 θ)) + C

Solution steps

\int sin^{2} (5 θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (10 θ)) 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 (10 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 (10 θ) d θ)

\int 1 d θ = θ

\int cos (10 θ) d θ = \frac{1}{10} sin (10 θ)

= \frac{1}{2} (θ - \frac{1}{10} sin (10 θ))

Add a constant to the solution

= \frac{1}{2} (θ - \frac{1}{10} sin (10 θ)) + C

\int \frac{x ^{2} - 5 x + 3}{x - 7} d x

x^{2} + y^{2} = x y y^{^{'}}

e^{- 2 s} (\frac{1}{4 s})

\int x (4 - x^{2})^{\frac{1}{2}} d x

\frac{\partial}{\partial y} (x y e^{x y})

What is the integral of sin^2(5θ) ?
The integral of sin^2(5θ) is 1/2 (θ-1/10 sin(10θ))+C