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Evaluate $$\int\frac{dx}{1+x^2}$$

Please help me find my mistake. I have integrated $\frac {1}{1+x^2}$ and gotten the correct result by making a mistake in the substitution. I imagined a triangle, with $1 = \cos\theta$ and $x = \sin\theta$ I then integrated $d\theta$ and got $\arctan$ a numerical result as this was a definite integral. While the result was correct, I realized that I should have substituted $dx$ for $\cos \theta$ but if I do this I get the wrong result.

I think I am making a mistake in the substitution thinking. Thank you!

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    $\begingroup$ Recall $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ $\endgroup$
    – Naren
    Jul 28, 2020 at 6:05
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    $\begingroup$ Re " $1 = \cos\theta$ and $x = \sin\theta$ " if $\cos\theta =1$ won't $\sin\theta=0$? $\endgroup$ Jul 28, 2020 at 6:07
  • $\begingroup$ @NarenNaruto Thanks, I know the derivative, but I wanted to solve it using trig substitution $\endgroup$
    – oliver
    Jul 28, 2020 at 6:09
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    $\begingroup$ @Oliver, Why dont you try with $x=\tan y$ ? $\endgroup$
    – Naren
    Jul 28, 2020 at 6:10
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    $\begingroup$ You've written sine and cosine when you mean the opposite leg and the adjacent leg, respectively. Don't get these mixed up (sine/cosine are ratios, not lengths) $\endgroup$ Jul 28, 2020 at 6:15

1 Answer 1

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The folly:

When you put $1=\cos\theta$, you are fixing the values of $\theta$ to be equal to $2n\pi+\frac{\pi}{2}$, where $n$ is an integer.


Fixing it:

If you do know the following, well and good, else another method follows:

$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$

The other method:

Substitute $x=\tan\theta$. Or, $dx=\sec^2\theta\cdot d\theta$. So the integral becomes: $$I=\int\frac{dx}{1+x^2}=\int\frac{\sec^2\theta\cdot d\theta}{1+\tan^2\theta}$$ Hence, $$I=\int\frac{\sec^2\theta}{\sec^2\theta}d\theta=\int d\theta$$ Can you finish?

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  • $\begingroup$ Hi, thanks so much. I can finish, as this is what I originally got to get the correct result. But I am not quite understanding why this would be case with cosine. If I have a triangle, with the hypothenuse being length $\sqrt{1+x^2}$ and the lower side 1 and the opposite x. Does theta have to be fixed? $\endgroup$
    – oliver
    Jul 28, 2020 at 6:15
  • $\begingroup$ Hi, thank you very much, thanks to a commenter above, I see how the substitution is not working! And how your comment above makes sense too now as this triangle doesn't exist with this theta. $\endgroup$
    – oliver
    Jul 28, 2020 at 6:18
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    $\begingroup$ @oliver: If you substitute $1=\cos\theta$ then $\theta$ is fixed; but according to this comment if yours, $\cos\theta$ happens to be $\frac{1}{\sqrt{1+x^2}}$ and now it is not fixed, and is a valid substitution, though it's usefulness is debatable ;) $\endgroup$ Jul 28, 2020 at 6:19

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