What is the integral of 4sin^2(θ) ?

The integral of 4sin^2(θ) is 2θ-sin(2θ)+C

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integral of 4sin^2(θ)

\int 4 sin^{2} (θ) d θ

2 θ - sin (2 θ) + C

Solution steps

\int 4 sin^{2} (θ) d θ

Take the constant out:

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

= 4 \cdot \int sin^{2} (θ) d θ

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Simplify

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

= 2 θ - sin (2 θ)

Add a constant to the solution

= 2 θ - sin (2 θ) + C

\int (x^{4} + 2 x^{3} - 5 x + 9) d x

x \to o lim (e^{x})

arcsin (8 x) + arccos (8 x)

x \to + 0 lim (\frac{sin ( 5 x )}{5})

n = 1 \sum \infty \frac{x ^{n}}{n 5 ^{n}}