What is the integral of 1/b sin(bx) ?

The integral of 1/b sin(bx) is -1/(b^2)cos(bx)+C

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integral of 1/b sin(bx)

\int \frac{1}{b} sin (b x) d x

- \frac{1}{b ^{2}} cos (b x) + C

Solution steps

\int \frac{1}{b} sin (b x) d x

Take the constant out:

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

= \frac{1}{b} \cdot \int sin (b x) d x

Apply u-substitution

: \int \frac{sin ( u )}{b} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = b x

= \frac{1}{b} \cdot \frac{1}{b} (- cos (b x))

Simplify

\frac{1}{b} \cdot \frac{1}{b} (- cos (b x)) : - \frac{1}{b ^{2}} cos (b x)

= - \frac{1}{b ^{2}} cos (b x)

Add a constant to the solution

= - \frac{1}{b ^{2}} cos (b x) + C

\int (2 x - 1)^{6} d x

\int \frac{2 ln ( x )}{x ^{2}} d x

y = x^{2} - 9 x, (7, - 14)

\int_{- 1}^{1} 3 x d x

x \to 0 + lim (x^{15} ln (x))

What is the integral of 1/b sin(bx) ?
The integral of 1/b sin(bx) is -1/(b^2)cos(bx)+C