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

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

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integral of-sin(ln(x))ln(x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

x \cdot \frac{d y}{d x} + 4 y = 1

x \frac{d y}{d x} + \frac{2 x + 1}{x + 1} y = x - 1

\frac{1}{1 - \frac{1}{x}}, 0

\int \frac{1}{( b ^{2} - x ^{2} ) ^{\frac{3}{2}}} d x

f (x) = 10 x^{2} + 10 x + 8 + ln (x)

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