What is the integral of x^3sin(ln(x^2)) ?

The integral of x^3sin(ln(x^2)) is -1/10 x^4cos(2ln(x))+1/5 x^4sin(2ln(x))+C

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integral of x^3sin(ln(x^2))

\int x^{3} sin (ln (x^{2})) d x

- \frac{1}{10} x^{4} cos (2 ln (x)) + \frac{1}{5} x^{4} sin (2 ln (x)) + C

Solution steps

\int x^{3} sin (ln (x^{2})) d x

Simplify

ln (x^{2}) : 2 ln (x)

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

Apply u-substitution

: \int e^{4 u} sin (2 u) d u

= \int e^{4 u} sin (2 u) d u

Apply Integration By Parts

: - \frac{1}{2} e^{4 u} cos (2 u) - \int - 2 e^{4 u} cos (2 u) d u

= - \frac{1}{2} e^{4 u} cos (2 u) - \int - 2 e^{4 u} cos (2 u) d u

Take the constant out:

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

= - \frac{1}{2} e^{4 u} cos (2 u) - (- 2 \cdot \int e^{4 u} cos (2 u) d u)

Apply Integration By Parts

: \frac{1}{2} e^{4 u} sin (2 u) - \int 2 e^{4 u} sin (2 u) d u

= - \frac{1}{2} e^{4 u} cos (2 u) - (- 2 (\frac{1}{2} e^{4 u} sin (2 u) - \int 2 e^{4 u} sin (2 u) d u))

Take the constant out:

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

= - \frac{1}{2} e^{4 u} cos (2 u) - (- 2 (\frac{1}{2} e^{4 u} sin (2 u) - 2 \cdot \int e^{4 u} sin (2 u) d u))

Therefore

\int e^{4 u} sin (2 u) d u = - \frac{1}{2} e^{4 u} cos (2 u) - (- 2 (\frac{1}{2} e^{4 u} sin (2 u) - 2 \cdot \int e^{4 u} sin (2 u) d u))

Isolate

\int e^{4 u} sin (2 u) d u

= - \frac{e ^{4 u} cos ( 2 u )}{10} + \frac{e ^{4 u} sin ( 2 u )}{5}

Substitute back

u = ln (x)

= - \frac{e ^{4 l n (x)} cos ( 2 ln ( x ) )}{10} + \frac{e ^{4 l n (x)} sin ( 2 ln ( x ) )}{5}

Simplify

- \frac{e ^{4 l n (x)} cos ( 2 ln ( x ) )}{10} + \frac{e ^{4 l n (x)} sin ( 2 ln ( x ) )}{5} : - \frac{1}{10} x^{4} cos (2 ln (x)) + \frac{1}{5} x^{4} sin (2 ln (x))

= - \frac{1}{10} x^{4} cos (2 ln (x)) + \frac{1}{5} x^{4} sin (2 ln (x))

Add a constant to the solution

= - \frac{1}{10} x^{4} cos (2 ln (x)) + \frac{1}{5} x^{4} sin (2 ln (x)) + C

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What is the integral of x^3sin(ln(x^2)) ?
The integral of x^3sin(ln(x^2)) is -1/10 x^4cos(2ln(x))+1/5 x^4sin(2ln(x))+C