What is the integral of 9(sin(2x))/(sin(x)) ?

The integral of 9(sin(2x))/(sin(x)) is 18sin(x)+C

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integral of 9(sin(2x))/(sin(x))

\int 9 \frac{sin ( 2 x )}{sin ( x )} d x

18 sin (x) + C

Solution steps

\int 9 \cdot \frac{sin ( 2 x )}{sin ( x )} d x

Take the constant out:

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

= 9 \cdot \int \frac{sin ( 2 x )}{sin ( x )} d x

Rewrite using trig identities

= 9 \cdot \int 2 cos (x) d x

Take the constant out:

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

= 9 \cdot 2 \cdot \int cos (x) d x

Use the common integral:

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

= 9 \cdot 2 sin (x)

Simplify

= 18 sin (x)

Add a constant to the solution

= 18 sin (x) + C

\frac{d}{d x} (\frac{π x}{2})

\frac{\partial}{\partial x} (\frac{1 + x}{1 + y})

\int 12 - x^{2} d x

7 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0

x^{2} \frac{d y}{d x} = \frac{4 x ^{2} - x - 2}{( x + 1 ) ( y + 1 )}

What is the integral of 9(sin(2x))/(sin(x)) ?
The integral of 9(sin(2x))/(sin(x)) is 18sin(x)+C