What is the integral of 7sin(ln(x)) ?

The integral of 7sin(ln(x)) is 7(1/2 xsin(ln(x))-1/2 xcos(ln(x)))+C

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integral of 7sin(ln(x))

\int 7 sin (ln (x)) d x

7 (\frac{1}{2} x sin (ln (x)) - \frac{1}{2} x cos (ln (x))) + C

Solution steps

\int 7 sin (ln (x)) d x

Take the constant out:

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

= 7 \cdot \int sin (ln (x)) d x

Apply Integration By Parts

: x sin (ln (x)) - \int cos (ln (x)) d x

= 7 (x sin (ln (x)) - \int cos (ln (x)) d x)

\int cos (ln (x)) d x = \frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))

= 7 (x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))))

Expand

x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))) : \frac{1}{2} x sin (ln (x)) - \frac{1}{2} x cos (ln (x))

= 7 (\frac{1}{2} x sin (ln (x)) - \frac{1}{2} x cos (ln (x)))

Add a constant to the solution

= 7 (\frac{1}{2} x sin (ln (x)) - \frac{1}{2} x cos (ln (x))) + C

x \to 0 lim (\frac{1}{1 + x})

\int sin^{6} (x) d x

x \to 0 lim (\frac{1}{4 x ^{2}})

f (x) = 1.8 sin (x) - x

\int \frac{15}{( x - 1 ) ( x ^{2} + 4 )} d x

What is the integral of 7sin(ln(x)) ?
The integral of 7sin(ln(x)) is 7(1/2 xsin(ln(x))-1/2 xcos(ln(x)))+C