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I have been taught that the indefinite integration of $$\int\frac{f'(x)}{f(x)}\,dx = \log |f(x)|$$

But this question was asked in AIEEE 2004:

$$\int\frac{dx}{\cos x - \sin x}$$

We may easily get that the answer is $$\text{My answer: }\frac{1}{\sqrt{2}}\log \left| \tan \left(\frac{\pi}{8}-\frac{x}{2}\right)\right| +c$$

But the options are

$$(i)\ \ \ \frac{1}{\sqrt{2}}\log \left| \tan \left(\frac{x}{2}-\frac{\pi}{8}\right)\right| +c$$ and $$(ii)\ \ \ \frac{1}{\sqrt{2}}\log \left| \tan \left(\frac{x}{2}+\frac{3\pi}{8}\right)\right| +c$$

and both are correct according to me. Am I missing something? What have I done wrong?

I am even not able to derive the second option from the first but you may differentiate both my answer and the last option to get the answer. You wont get the answer by differentiating the first option but theoretically the first option is the same as my answer.

EDIT More explanation

The differentiation of $$\frac{d}{dx}\left(\log\left|\tan \frac{x}{2}\right|\right) = \frac{1}{\sin x}$$

So option (i)'s differentiation is $$\frac{1}{\sin x - \cos x}$$

While my answer's and option (ii)'s differentiation is $$\frac{1}{\cos x - \sin x}$$

Megh Parikh
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3 Answers3

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Your answer and option 1 both are equal and incorrect while only option 2 is correct. I think you have mistaken and assumed that log|tanx|= log|cotx| which is incorrect and the correct one is log|tanx|= -log|cotx|.

ANANT
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The option 2 is correct ; option 1 and your answer is same and further, you have done wrong differentiation for your answer.

Purvam Modi
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Well, in the two options, the arguments to tan inside the log differ by $\pi/2$; that makes them rather different. I like the first one (assuming your earlier work is right -- seems to be on a quick reading), because by $\tan(-u) = -\tan(u)$, you can make it equal to your answer.

John Hughes
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  • You may differentiate both my answer and the last option to get the answer. You wont get the answer by differntiating the first option but theoretically the first option is the same as my answer. – Megh Parikh Feb 06 '15 at 12:40
  • Your answer and option 'i' are equal as functions, because $\tan$ is odd. Perhaps your differentiation work is incorrect, because equal functions have equal derivatives. Why don't you show us your work rather than explaining why we're wrong? – John Hughes Feb 06 '15 at 13:19
  • I have edited. Do you need more explanation? – Megh Parikh Feb 06 '15 at 13:32
  • I have edited more – Megh Parikh Feb 06 '15 at 13:37
  • Your answer has an absolute value in it; the differentiation that you offered does not. Do you think that the loss of an absolute value might explain a missing minus sign? – John Hughes Feb 06 '15 at 13:45
  • Maybe yes. But then I know only one rule and I dont know when not to use the rule. If you know more about it or a good resource can you link it please – Megh Parikh Feb 06 '15 at 13:54
  • Calculus, by Michael Spivak, the chapter called "techniques of integration" and its associated problems. Rather than give a bunch of rules, he carefully presents the theorems that justify the rules. The rule you cite, in computing the derivative above, for instance, has a constraint: it's only valid where $\tan(\frac{x}{2})$ is positive; in applying it, you ignored that constraint (perhaps because you didn't know it was there!) and therefore ended up with an error. – John Hughes Feb 06 '15 at 14:10
  • In indefinite integration we cant know if its positive or negative – Megh Parikh Feb 06 '15 at 14:28
  • I believe you need to study carefully just what an indefinite integral is, including its domain of definition. – John Hughes Feb 06 '15 at 15:50