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

The integral of (sin^7(ln(x)))/x is -cos(ln(x))+cos^3(ln(x))-(3cos^5(ln(x)))/5+(cos^7(ln(x)))/7+C

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integral of (sin^7(ln(x)))/x

\int \frac{sin ^{7} ( ln ( x ) )}{x} d x

- cos (ln (x)) + cos^{3} (ln (x)) - \frac{3 cos ^{5} ( ln ( x ) )}{5} + \frac{cos ^{7} ( ln ( x ) )}{7} + C

Solution steps

\int \frac{sin ^{7} ( ln ( x ) )}{x} d x

Apply u-substitution

: \int sin^{7} (u) d u

= \int sin^{7} (u) d u

Rewrite using trig identities

= \int (1 - cos^{2} (u))^{3} sin (u) d u

Apply u-substitution

: \int - (1 - v^{2})^{3} d v

= \int - (1 - v^{2})^{3} d v

Expand

- (1 - v^{2})^{3} : - 1 + 3 v^{2} - 3 v^{4} + v^{6}

= \int - 1 + 3 v^{2} - 3 v^{4} + v^{6} d v

Apply the Sum Rule:

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

= - \int 1 d v + \int 3 v^{2} d v - \int 3 v^{4} d v + \int v^{6} d v

\int 1 d v = v

\int 3 v^{2} d v = v^{3}

\int 3 v^{4} d v = \frac{3 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= - v + v^{3} - \frac{3 v ^{5}}{5} + \frac{v ^{7}}{7}

Substitute back

= - cos (ln (x)) + cos^{3} (ln (x)) - \frac{3 cos ^{5} ( ln ( x ) )}{5} + \frac{cos ^{7} ( ln ( x ) )}{7}

Add a constant to the solution

= - cos (ln (x)) + cos^{3} (ln (x)) - \frac{3 cos ^{5} ( ln ( x ) )}{5} + \frac{cos ^{7} ( ln ( x ) )}{7} + C

\frac{1}{x ^{8}} y^{^{'}} = 12 + 15 y

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

y = x^{3}, at x = - 9

(1 + sin (x))^{- 4}

\int (\frac{1}{x} + \frac{1}{3 - x}) d x

What is the integral of (sin^7(ln(x)))/x ?
The integral of (sin^7(ln(x)))/x is -cos(ln(x))+cos^3(ln(x))-(3cos^5(ln(x)))/5+(cos^7(ln(x)))/7+C