What is the integral of 1in(x) ?

The integral of 1in(x) is -dcos(x)+C

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integral of 1in(x)

\int 1 d sin (x) d x

- d cos (x) + C

Solution steps

\int 1 \cdot d sin (x) d x

Take the constant out:

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

= 1 \cdot d \cdot \int sin (x) d x

Simplify

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

Use the common integral:

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

= d (- cos (x))

Simplify

= - d cos (x)

Add a constant to the solution

= - d cos (x) + C