What is the integral of 3sin(2θ) ?

The integral of 3sin(2θ) is -3/2 cos(2θ)+C

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integral of 3sin(2θ)

\int 3 sin (2 θ) d θ

- \frac{3}{2} cos (2 θ) + C

Solution steps

\int 3 sin (2 θ) d θ

Take the constant out:

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

= 3 \cdot \int sin (2 θ) d θ

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 θ

= 3 \cdot \frac{1}{2} (- cos (2 θ))

Simplify

3 \cdot \frac{1}{2} (- cos (2 θ)) : - \frac{3}{2} cos (2 θ)

= - \frac{3}{2} cos (2 θ)

Add a constant to the solution

= - \frac{3}{2} cos (2 θ) + C

3 x^{4} + 9 x^{3} + 11

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

y^{^{''}} - 6 y^{^{'}} + 13 = 0

(- 4, 2), (1, - 4)

\frac{d y}{d x} - 3 y = 0