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Can anyone help me figure out what is wrong in the following step:

$$\begin{align}\int \tan x\ dx &= \int (\sec x \sin x)\ dx\\ &= -\sec x \cos x + \int \sec x\tan x\cos x\ dx\\ &= -1 + \int \tan x dx\end{align}$$

So I got $\displaystyle\int \tan x\ dx = -1 + \int \tan x\ dx$ that is $0 = -1$ ?

Any help will be really appreciated!

Now I understand. So an indefinite integral can be defined only upto a constant. Thanks a lot!

Tunk-Fey
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Jenny
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2 Answers2

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What you've lost is constant of integration of $\sin x$. If instead of $-\cos x$ you take $C-\cos x$, then you'll have:

$$\int \tan x\;dx=\int(\sec x\sin x)\;dx=\\ =-\sec x\cos x+C\sec x+\int(\cos x-C)\tan x\sec x\;dx=\\ =-1+C\sec x+\int \tan x\;dx-C\int \tan x\sec x\;dx.$$

Then instead of $-1=0$ you'll have

$$C\left(\sec x-\int \tan x\sec x\;dx\right)=1,$$

which does make sense.

Ruslan
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Think it in this way, say you're evaluating the definite integral $$\int_{a}^b \tan x dx$$ Then according to your steps you'll get $$\int_{a}^b\tan x \ dx=[-1]_a^b+\int_{a}^b \tan x\ dx=\int_{a}^b\tan x \ dx$$So nothing gets changed.

Samrat Mukhopadhyay
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    I think it would be more instructive to show how exactly losing constant of integration in indefinite integral leads to the nonsensical result in OP. Your answer just hides the problem instead of making it understandable. – Ruslan May 23 '14 at 08:42