Let $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\tan^{-1}(x^2)+\cot^{-1}(x^2)$. I need to find the primitive function of $f$.
Now, I know that f is a constant function, because its $f'(x)=0$. And I think that let's say if $F:\mathbb{R}\rightarrow\mathbb{R}$ is its primitive then $F'(x)=f(x)$ which means that $F'(x)$ needs also to be a constant. I think I managed to solve it $\int \tan^{-1}(x^2)+\cot^{-1}(x^2)dx=\int \frac{\pi}{2}dx=\frac{\pi}{2}x+C$, am I right?
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Ghost
- 1,105
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$$\arctan(x^2)$$ or $$(\arctan(x))^2$$ – Dr. Sonnhard Graubner Sep 19 '17 at 18:51
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1Try evaluating $f$ at some convenient $x$ value to see what the constant is. – Robert Israel Sep 19 '17 at 18:52
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@Dr.SonnhardGraubner $arctan(x^2)$ – Ghost Sep 19 '17 at 18:52
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@RobertIsrael And what I do after? $f(x)=\frac{\pi}{2}$. – Ghost Sep 19 '17 at 18:53
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try Integration by parts – Dr. Sonnhard Graubner Sep 19 '17 at 18:55
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Hint: $$\arctan(x^2) = \frac \pi 2 - \operatorname{arccot}(x^2)$$ – George C Sep 19 '17 at 18:55
1 Answers
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Yes, you are correct.
From,
$$\tan(x^2) = \operatorname{cot}\left(\frac \pi 2 - x^2\right)$$
We have,
$$\arctan(x^2) = \frac \pi 2 - \operatorname{arccot}(x^2)$$
Hence,
$$\int (\arctan(x^2) + \operatorname{arccot}(x^2)) \mathrm dx = \int \left(\frac \pi 2 - \operatorname{arccot}(x^2) + \operatorname{arccot}(x^2)\right)\mathrm dx $$
Which is quite easy to evaluate,
$$\int \frac \pi 2 \mathrm dx = \frac \pi 2 x + C$$
Proof of first line
$$\operatorname{cot}\left(\frac \pi 2 - x^2\right) = \frac{\cos\left(\frac \pi 2 - x^2\right)}{\sin\left(\frac \pi 2 - x^2\right)} = \frac{\sin(x^2)}{\cos(x^2)} = \tan(x^2)$$
George C
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A little technicality: your first implication is not necessarily right because the functions $\tan$ and $\cot$ are not invertible and you have to choose a branch. In this case, with the usual conventions, you are safe. – Sep 19 '17 at 19:19
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You're right, but my use of $\operatorname{arccot}$ and $\arctan$ should make clear what branch was chosen. Your deleted answer did remind of me of the discontinuity at $x = 0$ that I stupidly missed, I'll address that now. – George C Sep 19 '17 at 19:26
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I deleted after noticing that the argument is always positive :-) But I wanted to remind that identities like $\tan a=\tan b$ can't be blindly turned to $a=b$ nor $\arctan a=\arctan b$. – Sep 19 '17 at 19:36
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Ah, I wasn't thinking of that at the time, I was more thinking of substituting a suitable $x^2$ to make sure that the argument of $\tan$/$\operatorname{cot}$ lied in the "principal range". Odd because I'm normally the one to pick up on said technicalities. Thanks for the shout :) – George C Sep 19 '17 at 20:00