What is the antiderivative of #sin (lnx)#?

Question

Cameron Alexander · Accepted Answer

#(cos(lnx))/x# According to the chain rule, #d/dx[sin(u)]=cos(u)*(du)/dx#. Thus, #d/dx[sin(lnx)]=cos(lnx)*d/dx[lnx]# #=cos(lnx)*1/x=(cos(lnx))/x#

Cameron Alexander · Answer

#x/2(sin(ln(x))-cos(ln(x))) + C#

In the given context, finding the antiderivative of #sin(ln(x))# is equivalent to solving the indefinite integral #intsin(ln(x))dx#

To do so, we will make use of integration by substitution and integration by parts. We will need to do some additional tricks with these beyond the standard basic methods as well.

To begin, we let #I = intsin(ln(x))dx#

Substitution:

Let #t = ln(x) => dt = 1/x dx#

Unfortunately, this is not enough, as we do not have #1/xdx# in our integral. To fix this, note that from the substitution:

#e^t = e^ln(x) = x#

Then, multiplying both sides of our derivative above by #x#, we have

#dx = xdt = e^tdt#

Continuing with the substitution,

#I = intsin(t)e^tdt#

Integration by parts (i):

Let #u = sin(t)# and #dv = e^tdt# #=> du = cos(t)dt# and #v = e^t#

Applying the integration by parts formula #intudv = uv - intvdu#

#inte^tsin(t)dt = e^tsin(t) - inte^tcos(t)dt#

Integration by parts (ii):

Let #u = cos(t)# and #dv = e^tdt# #=> du = -sin(t)dt# and #v = e^t#

Again, applying the formula,

# inte^tcos(t)dt = e^tcos(t) - int(- e^tsin(t))dt#

Substituting this into the result of the first integration by parts,

#I = e^tsin(t) - e^tcos(t) - inte^tsin(t)dt#

# = e^t(sin(t)-cos(t)) - I#

#=> 2I = e^t(sin(t)-cos(t))#

#=> I = e^t/2(sin(t)-cos(t))#

substituting #t = ln(x)# back in, this gives

#I = x/2(sin(ln(x)) - cos(ln(x)))#

And since we lost the constant when we did the trick adding #I# to both sides, let's put it back in now to get the final result.

#I = x/2(sin(ln(x))-cos(ln(x))) + C#

Ezra Adams · Answer

undefined The antiderivative of $ \sin(\ln x) $ is not expressible in terms of elementary functions. It cannot be represented using combinations of basic functions such as polynomials, exponential functions, logarithmic functions, trigonometric functions, and their inverses. However, it can be expressed using special functions like the exponential integral function or the sine integral function, but these representations are more complex and not as commonly used. Therefore, the antiderivative of $ \sin(\ln x) $ is typically denoted by $ \text{Si}(x) $, the sine integral function, defined as:

$$ \text{Si}(x) = \int \frac{\sin t}{t} \, dt $$

But it's important to note that this integral is not expressible in terms of elementary functions.