What is the integral of sin(log_{10}(x)) 1/x ?

The integral of sin(log_{10}(x)) 1/x is -ln(10)cos(log_{10}(x))+C

integral of sin(log_{10}(x)) 1/x

\int sin (lo g_{10} (x)) \frac{1}{x} d x

- ln (10) cos (lo g_{10} (x)) + C

Solution steps

\int sin (lo g_{10} (x)) \frac{1}{x} d x

Apply u-substitution

= \int ln (10) sin (u) d u

Take the constant out:

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

= ln (10) \cdot \int sin (u) d u

Use the common integral:

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

= ln (10) (- cos (u))

Substitute back

u = lo g_{10} (x)

= ln (10) (- cos (lo g_{10} (x)))

Simplify

= - ln (10) cos (lo g_{10} (x))

Add a constant to the solution

= - ln (10) cos (lo g_{10} (x)) + C

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What is the integral of sin(log_{10}(x)) 1/x ?
The integral of sin(log_{10}(x)) 1/x is -ln(10)cos(log_{10}(x))+C