How do you integrate $\int sin (ln x)$ $int sin(lnx)$ using integration by parts?

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Answer 1 · 2015-11-03T22:07:46

$\frac{1}{2} ([x sin (ln (x))] - [x cos (ln (x))])$ $1/2([xsin(ln(x))]-[xcos(ln(x))])$

$\int sin (ln (x)) d x$ $intsin(ln(x))dx$

Let's $u = ln (x)$ $u = ln(x)$

$d u = \frac{d x}{x}$ $du = dx/x$

$d x = x d u$ $dx=xdu$

$x = e^{u}$ $x=e^u$

then

$\int e^{u} sin (u) d u$ $inte^usin(u)du$

By part : $d v = e^{u}; v = e^{u}; w = sin (u); d w = cos (u)$ $dv = e^u ; v = e^u ; w = sin(u) ; dw = cos(u)$

$[v \cdot w] - \int d w \cdot v$ $[v*w]-intdw*v$

$\int e^{u} sin (u) d u = [e^{u} sin (u)] - \int e^{u} cos (u) d u$ $inte^usin(u)du=[e^usin(u)]-inte^ucos(u)du$

By part again

$d v = e^{u}; v = e^{u}; w = cos (u); d w = - sin (u)$ $dv=e^u ; v = e^u ; w = cos(u) ; dw = -sin(u)$

$\int e^{u} sin (u) d u = [e^{u} sin (u)] - ([e^{u} cos (u)] + \int e^{u} sin (u) d u)$ $inte^usin(u)du=[e^usin(u)]-([e^ucos(u)]+inte^usin(u)du)$

$\int e^{u} sin (u) d u = [e^{u} sin (u)] - [e^{u} cos (u)] - \int e^{u} sin (u) d u$ $inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]-inte^usin(u)du$

$2 \int e^{u} sin (u) d u = [e^{u} sin (u)] - [e^{u} cos (u)]$ $2inte^usin(u)du=[e^usin(u)]-[e^ucos(u)]$

$\int e^{u} sin (u) d u = \frac{1}{2} ([e^{u} sin (u)] - [e^{u} cos (u)])$ $inte^usin(u)du=1/2([e^usin(u)]-[e^ucos(u)])$

Substitute back for $u = ln (x)$ $u = ln(x)$

$\int sin (ln (x)) d x = \frac{1}{2} ([x sin (ln (x))] - [x cos (ln (x))])$ $intsin(ln(x))dx=1/2([xsin(ln(x))]-[xcos(ln(x))])$

How do you integrate ∫sin(lnx)int sin(lnx) using integration by parts?