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

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

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integral of-1/b cos(bx)

\int - \frac{1}{b} cos (b x) d x

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

Solution steps

\int - \frac{1}{b} cos (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 cos (b x) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = b x

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

Simplify

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

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

Add a constant to the solution

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

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