What is the integral of 4sin(2θ) ?

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

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integral of 4sin(2θ)

\int 4 sin (2 θ) d θ

- 2 cos (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

Apply u-substitution

: \int sin (u) \frac{1}{2} d u

= 4 \cdot \int sin (u) \frac{1}{2} d u

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 sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= 4 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 θ

= 4 \cdot \frac{1}{2} (- cos (2 θ))

Simplify

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

= - 2 cos (2 θ)

Add a constant to the solution

= - 2 cos (2 θ) + C

\int_{- 2}^{2} x^{2} - 4 d x

\frac{d}{d x} (3 sin^{2} (x) + sec (2 x))

f (x) = x + \frac{1}{x}, at x = 4

\int (3 t + 5) cos (\frac{t}{4}) d t

f (x) = 3 x^{(2)} - 5 x - 6