What is the integral of (sin(ln(x))cos(2ln(x)))/(2x) ?

The integral of (sin(ln(x))cos(2ln(x)))/(2x) is 1/6 (3cos(ln(x))-2cos^3(ln(x)))+C

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integral of (sin(ln(x))cos(2ln(x)))/(2x)

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

\frac{1}{6} (3 cos (ln (x)) - 2 cos^{3} (ln (x))) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Expand

\frac{( 1 - 2 sin ^{2} ( ln ( x ) ) ) sin ( ln ( x ) )}{x} : \frac{sin ( ln ( x ) )}{x} - \frac{2 sin ^{3} ( ln ( x ) )}{x}

= \frac{1}{2} \cdot \int \frac{sin ( ln ( x ) )}{x} - \frac{2 sin ^{3} ( ln ( x ) )}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{sin ( ln ( x ) )}{x} d x - \int \frac{2 sin ^{3} ( ln ( x ) )}{x} d x)

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

\int \frac{2 sin ^{3} ( ln ( x ) )}{x} d x = 2 (- cos (ln (x)) + \frac{cos ^{3} ( ln ( x ) )}{3})

= \frac{1}{2} (- cos (ln (x)) - 2 (- cos (ln (x)) + \frac{cos ^{3} ( ln ( x ) )}{3}))

Simplify

\frac{1}{2} (- cos (ln (x)) - 2 (- cos (ln (x)) + \frac{cos ^{3} ( ln ( x ) )}{3})) : \frac{1}{6} (3 cos (ln (x)) - 2 cos^{3} (ln (x)))

= \frac{1}{6} (3 cos (ln (x)) - 2 cos^{3} (ln (x)))

Add a constant to the solution

= \frac{1}{6} (3 cos (ln (x)) - 2 cos^{3} (ln (x))) + C

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

4 y d x - x y - 3 x d y = 0

\frac{d}{d x} ((x + 2) (2 x^{2} - 3))

\frac{d}{d x} (9 x^{8})

x^{2} - 9

What is the integral of (sin(ln(x))cos(2ln(x)))/(2x) ?
The integral of (sin(ln(x))cos(2ln(x)))/(2x) is 1/6 (3cos(ln(x))-2cos^3(ln(x)))+C