My friend had the following doubt:

\( \int { \frac { dx }{ x\ln { x } } } \)

You might laugh at me and say this is a straightforward integration by substitution which gets us to

\( \ln { (\ln { x)+C } } \)

But what if we try integration by parts???

\( I=\int { \frac { dx }{ x\ln { x } } } =\frac { 1 }{ \ln { x } } \int { \frac { dx }{ x } } -\int { \left( -\frac { 1 }{ { (\ln { x } ) }^{ 2 } } \frac { 1 }{ x } \int { \frac { dx }{ x } } \right) dx } =\frac { 1 }{ \ln { x } } \ln { x } +\int { \frac { dx }{ x\ln { x } } } =1+I\quad \\ \Rightarrow I=I+1\Rightarrow 1=0 \)

Please tell me where are we wrong?

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## Comments

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TopNewestYes the approach using integration by parts tells us that something is wrong, but you will be surprised that both answers are true!

As long as we are working with indefinite integrals, the key here is the constant of integration. You forgot to add a constant when you integrated by parts. we know that this constant is arbitrary, that is, you could use \(C+1\) as constant of integration rather than \(C\).

Therefore this is correct using either methods, and you can assure that by two ideas :

1) you can differentiate both answers, you will get the same initial function.

2) suppose you are dealing with definite integral: when you use the fundamental theorem of calculus, the excess "\(1\)" will be cancelled.

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This is one of my favorite "fallacies" in calculus.

In a similar vein: How to show that \( \sin \theta = \cos \theta = 0 \).

\[ \int \sin \theta \cos \theta \, d \theta = \int \sin \theta \, d \sin \theta = \frac{1}{2} \sin ^2 \theta, \]

\[ \int \sin \theta \cos \theta \, d \theta = - \int \cos \theta \, d \cos \theta = - \frac{1}{2} \cos^2 \theta \]

Hence, \( \frac{1}{2} \sin^2 \theta = - \frac{1}{2} \cos ^2 \theta \). But since squares are non-negative, they must both be 0.

Of course, as pointed out by Hasan, the correct conclusion should be:

\[ \frac{1}{2} \sin ^2 \theta + C_1 = - \frac{1}{2} \cos ^2 \theta + C_2 \Rightarrow \sin^2 \theta + \cos^2 \theta = C_2 - C_1, \]

which is a well known fact.

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Yes, examples like these "fake" proofs are somehow interesting.

Another example is The integral \(\displaystyle \int \mathrm \tan x \sec^2 x \mathrm {d}x\) . Using substitution \(u=\sec x\) and \(u=\tan x\) will yield two obviously different answers.

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And wouldn't this imply that \(C_2-C_1=1\)?

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How is integration of \sin \theta \cos \theta =sin^2(theta) ... if we assume \sin \theta = t, then we will get t^2/2 ...thus (sin theta)^2 / 2 ...and how can u write dsin(theta)...I understood the fallacies, but plz respond to my doubts

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As explained by @John Muradeli , I used the substitution of \( u = \sin \theta \) and \( u = \cos \theta \) respectively. It is a shorthand way of expressing it, without stating it explicitly.

Note: I've edited your comment to show you how to use the brackets \ ( \ ) around the math code to make the formulas display correctly.

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Wolfram|Alpha agrees.

He just used the substitution you proposed, just in a disguised way. And you're right.Calvin made a boo-boo

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\sin^2{x}= \(\sin^2{x}\)

\theta = \(\theta\)

Use \(\(...\)) for latex. (... replace with the code, ex. sin{x})

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\(\int{tdt}\)

Now think of \(t\) as a blob of what it originally was (\(\sin{(x)}\)). If you do so, you end up with

\(\int{\sin{(x)}d(\sin{(x)}})\)

-see? Pretty clever, eh?

And yes you do end up with \(\frac{t^2}{2}+C\), or, \(\frac{1}{2}\sin^2{x}+C\).

P.S.

If you wonder why he did it, sometimes in problems its easy to see. If you notice, the stuff after \(d\) is just the stuff he took out - integrated. This is so that to undo the \(d"stuff"\), when you differentiate the "stuff" it will get back into the argument and \(d\) will take its original form. This is a bit a wacko of an explanation but you can probably see it better from the \(t\) substitution.

Hope this helped.

Cheers,

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Edit: Use \( \sin {x} \). You can edit this comment (click the pencil in the top right), to see the actual code used.

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Cool?

\pi =\(\pi\)

\infty = \(\infty\)

\Rightarrow = \(\Rightarrow\)

When raising to a power that includes more than two characters, encase the exponent with {...}. Ex.

\e^-x=\(e^-1x\). \e^{-x}=\(e^{-x}\)

P.S. - double enter to skip a line

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@hasan kassim @Krishna Jha Can you add this to the Wiki page of a suitable skill in Integration by Parts?

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How do I do that sir?

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@Krishna Jha Can you add this to Integration by Parts, so that others can be kept aware of this common mistake? Thanks!

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OK.. But how to do that sir? Is this fine??

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bro,, this is indefinite integral,, several functions infact infinite functions all differing by a constant give us the same derivative,,

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The difference between the primitives of a function is a constant.

Let \(F ' (x) = G ' (x) = f(x) \).

Then \(F(x) = G(x) + C\).

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You cannot take a number that needs to be integrated out of the integration

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U can't use integration by parts for integrating two same functions

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why not??

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