What is $\int sin (ln x) d x$ $int sin(lnx) dx$ ?

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Answer 1 · 2015-12-17T23:27:04

Use integration by parts twice and solve for the integral to find

$\int sin (ln (x)) d x = \frac{x (sin (ln (x)) - cos (ln (x)))}{2} + C$ $intsin(ln(x))dx = (x(sin(ln(x)) - cos(ln(x))))/2 + C$

We will proceed through integration by parts.

Let $I = \int sin (ln (x)) d x$ $I = intsin(ln(x))dx$

Integration by Parts (1)

Let $u = sin (ln (x))$ $u = sin(ln(x))$ and $d v = d x$ $dv = dx$

Then $d u = \frac{cos (ln (x))}{x} d x$ $du = cos(ln(x))/xdx$ and $v = x$ $v = x$

Applying the integration by parts formula $\int u d v = u v - \int v d u$ $intudv = uv - intvdu$

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

$= x sin (ln (x)) - \int cos (ln (x)) d x$ $= xsin(ln(x)) - intcos(ln(x))dx$

Integration by Parts (2)

Let $u = cos (ln (x))$ $u = cos(ln(x))$ and $d v = d x$ $dv = dx$

Then $d u = - \frac{sin (ln (x))}{x}$ $du = -sin(ln(x))/x$ and $v = x$ $v = x$

Applying the formula, we have

$\int cos (ln (x)) d x = x cos (ln (x)) - \int - sin (ln (x)) d x$ $intcos(ln(x))dx = xcos(ln(x)) - int-sin(ln(x))dx$

$= x cos (ln (x)) + \int sin (ln (x)) d x$ $= xcos(ln(x)) + intsin(ln(x))dx$

$= x cos (ln (x)) + I$ $= xcos(ln(x)) + I$

Substituting this into the result from the first integration by parts, we obtain

$I = x sin (ln (x)) - x cos (ln (x)) - I$ $I = xsin(ln(x)) - xcos(ln(x)) - I$

$\Rightarrow 2 I = x (sin (ln (x)) - cos (ln (x)))$ $=> 2I = x(sin(ln(x)) - cos(ln(x)))$

$\Rightarrow I = \frac{x (sin (ln (x)) - cos (ln (x)))}{2}$ $=> I = (x(sin(ln(x)) - cos(ln(x))))/2$

But in the process of adding $I$ $I$ to both sides, we lost the constant, and so for our final answer, we add it back in to get

$I = \int sin (ln (x)) d x = \frac{x (sin (ln (x)) - cos (ln (x)))}{2} + C$ $I = intsin(ln(x))dx = (x(sin(ln(x)) - cos(ln(x))))/2 + C$

What is int sin(lnx) dx∫sin(lnx)dxint sin(lnx) dx?