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### Integration by Parts

```
Date: 01/07/97 at 21:57:35
From: Agustinus
Subject: Integral problems

I'm stuck - my homework is to find the result of:

integral sin (ln x) dx.

I have done several things but I don't think they worked.
Here's what I did:

let v  = ln x
dv = 1/x dx
dx = x dv

So the integral becomes:

integral x * sin v dv

The result is:

-x * cos v + c
-x * cos ln x + c

This is incorrect, isn't it? (It's too easy.)
Thanks.
```

```
Date: 01/08/97 at 03:42:36
From: Doctor Pete
Subject: Re: Integral problems

Hi,

I will point out what is incorrect with your approach.

Note that if you were to use the substitution:

v = Log[x] (what you write as "ln x")
dv = 1/x dx,

then:

Int[Sin[Log[x]] dx] = Int[Sin[v]x dv]

But if you were to say:

Int[Sin[v]x dv] = -x Cos[v] + C
= -x Cos[Log[x]] + C,

this is not right, because the substitution means that x is not a
constant!  In particular, v and x are related to each other. When we
take the integral of a function, we take it with respect to some
variable, which is indicated by dv or dx or d(something). That is:

Int[Sin[v]x dv]

is an integral with respect to the variable v. But v depends on x;
thus we cannot treat either v or x as constants. In order to avoid
this confusion, we need to do the substitution in full; that is, we
should have written:

v = Log[x]
x = Exp[v]
dx = Exp[v] dv

Hence:

Int[Sin[Log[x]] dx] = Int[Exp[v]Sin[v] dv].

Now, we may take the integral of the righthand side, because the
entire integrand (function to be integrated) is in terms of the
variable v.  To do this, however, is another matter entirely; it
requires integration by parts.  By selecting:

p = Sin[v]
dp = Cos[v] dv
dq = Exp[v] dv
q = Exp[v]

so Int[p dq] = pq - Int[q dp], or:

Int[Exp[v]Sin[v] dv] = Exp[v]Sin[v] - Int[Exp[v]Cos[v] dv]

To integrate the integral on the right, we do a similar substitution,
this time with:

p = Cos[v]
dp = -Sin[v] dv
dq = Exp[v] dv
q = Exp[v]

which gives:

Int[Exp[v]Cos[v] dv] = Exp[v]Cos[v] + Int[Exp[v]Sin[v] dv]

Now, you might think that we have just gone around in a circle, but
this is not the case, for if you substitute this second equation into
the first, you will find that the Int[Exp[v]Sin[v] dv] does not
cancel, and you will also find that you now have an equation which
says that twice this integral is equal to:

Exp[v]Sin[v] - Exp[v]Cos[v]

Finally, invert the substitution of v for x to obtain an integral in
terms of the variable x.

All this can be simplified considerably if you notice that the
original integral:

Int[Sin[Log[x]] dx]

is directly integrable by parts, with the substitution:

p = Sin[Log[x]]
dp = Cos[Log[x]]/x
dq = dx
q = x

I will leave it as an excercise for you to carry out this alternative,
and shorter, computation.  It is fairly straightfoward if you
understand the first method.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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