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

The integral of sin^2(ln(x)) is 1/10 (5x-2xsin(2ln(x))-xcos(2ln(x)))+C

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integral of sin^2(ln(x))

\int sin^{2} (ln (x)) d x

\frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x))) + C

Solution steps

\int sin^{2} (ln (x)) d x

Rewrite using trig identities

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

\int 1 d x = x

\int cos (2 ln (x)) d x = \frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))

= \frac{1}{2} (x - (\frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))))

Simplify

\frac{1}{2} (x - (\frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x)))) : \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x)))

= \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x)))

Add a constant to the solution

= \frac{1}{10} (5 x - 2 x sin (2 ln (x)) - x cos (2 ln (x))) + C

\int \frac{2 x - 9}{x ^{2} - 9 x + 1} d x

\frac{\partial}{\partial x} (\frac{x y}{( x ^{2} - y ^{2} ) ^{2}})

y = x^{2} + 1, (- 2, 5)

x, at 25

x^{4} - 2 x^{2} + 4

What is the integral of sin^2(ln(x)) ?
The integral of sin^2(ln(x)) is 1/10 (5x-2xsin(2ln(x))-xcos(2ln(x)))+C