What is the integral of (3cos(2θ)) ?

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

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integral of (3cos(2θ))

\int (3 cos (2 θ)) d θ

\frac{3}{2} sin (2 θ) + C

Solution steps

\int 3 cos (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 cos (2 θ) d θ

Apply u-substitution

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

= 3 \cdot \int cos (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 cos (u) d u

Use the common integral:

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

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

Substitute back

u = 2 θ

= 3 \cdot \frac{1}{2} sin (2 θ)

Simplify

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

= \frac{3}{2} sin (2 θ)

Add a constant to the solution

= \frac{3}{2} sin (2 θ) + C

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

x^{2}, - x + 5, 0

\frac{8 x ^{3} - 2 x + 9}{x}

\frac{\partial}{\partial u} (\frac{u}{u + v})

(x - 2) (3 x + 3)

What is the integral of (3cos(2θ)) ?
The integral of (3cos(2θ)) is 3/2 sin(2θ)+C