5

$$ \int_0^1\left[\left(\tan ^{-1} x\right)^2-\frac{\ln \left(1+x^2\right)}{1+x^2}\right] d x $$

What I did :

Substitute$$ x=\tan \theta $$

$$ \int_0^{\pi / 4}(\theta)^2 \sec ^2 \theta -\ln \left(\sec ^2 \theta\right) d \theta $$

But now I can't integrate the first term. Thanks in advance.

MathFail
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SavageTomato
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6 Answers6

5

Integrate both terms by parts separately

$$$$

$$I = \int_0^1 (\tan^{-1}x)^2\:dx = x(\tan^{-1} x)^2\Bigr|_0^1-\int_0^1\frac{2x\tan^{-1}x}{1+x^2}\:dx$$

$$$$

$$J = \int_0^1 \frac{\log(1+x^2)}{1+x^2}\:dx = \log(1+x^2)\tan^{-1}x\Bigr|_0^1 - \int_0^1 \frac{2x\tan^{-1}x}{1+x^2}\:dx$$

$$$$

The integral we want is $I-J$ and both integral terms cancel, leaving us with

$$$$

$$I-J = \left[x\tan^{-1}x-\log(1+x^2)\right]\tan^{-1}x\Bigr|_0^1 = \boxed{\frac{\pi^2}{16} - \frac{\pi\log 2}{4}}$$

Ninad Munshi
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3

Probably a non-intuitive way of solving this, so feel free to downvote and comment if it doesn't help:

$$ \begin{align} I &:= \int_{0}^{1}\left(\arctan\left(x\right)^{2}-\frac{\ln\left(1+x^{2}\right)}{1+x^{2}}\right)dx \\ &= \int_{0}^{1}\left(\arctan\left(x\right)^{2}+\frac{2x\arctan\left(x\right)}{1+x^{2}}-\frac{2x\arctan\left(x\right)}{1+x^{2}}-\frac{\ln\left(1+x^{2}\right)}{1+x^{2}}\right)dx \\ &= \int_{0}^{1}\left(\frac{d}{dx}x\arctan\left(x\right)^{2}-\frac{d}{dx}\ln\left(1+x^{2}\right)\arctan\left(x\right)\right)dx \\ &= \int_{0}^{1}\frac{d}{dx}x\arctan\left(x\right)^{2}dx-\int_{0}^{1}\frac{d}{dx}\ln\left(1+x^{2}\right)\arctan\left(x\right)dx \\ &= \left[x\arctan\left(x\right)^{2}\right]_{0}^{1}-\left[\ln\left(1+x^{2}\right)\arctan\left(x\right)\right]_{0}^{1} \\ &= \frac{\pi^{2}}{16}-\frac{\pi\ln\left(2\right)}{4}. \\ \end{align} $$

Accelerator
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2

Hint: \begin{align*} \int\arctan^2(x)\mathrm dx&=x\arctan^2(x)-\int\frac{2x}{1+x^2}\arctan(x)\mathrm dx\\ &=x\arctan^2(x)-\log(1+x^2)\arctan(x)+\int\frac{\log(1+x^2)}{1+x^2}\mathrm dx \end{align*}

Mostafa
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1

$$\begin{align} \int_0^{\pi/4} x^2\sec^2x dx&=\int_0^{\pi/4} 2x~d\tan x\\ \\ &=\frac{\pi^2}{16} -\int_0^{\pi/4} 2x\tan x~dx\\ \\ &=\frac{\pi^2}{16} +\int_0^{\pi/4} 2x~d\ln\cos x\\ \\ &=\frac{\pi^2}{16} +\frac{\pi}2\ln\frac{\sqrt2}2-2\int_0^{\pi/4} \ln\cos x~dx\\ \\ &=\frac{\pi^2}{16} -\frac{\pi}4\ln2-G+\frac{\pi}2\ln2\\ \\ &=\frac{\pi^2}{16} +\frac{\pi}4\ln2-G\end{align}$$

The integral $\int_0^{\pi/4} \ln\cos x~dx$ can be found here.

MathFail
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  • why downvote??? – MathFail Jun 01 '23 at 18:09
  • I didn't downvote or upvote because I thought the computation to Catalan constant was nice. But at the same time, it's not really relevant to OP's problem if this is from a list of simple calculus questions as they claim, and misleading to think that appeals to special functions or constants are necessary. So, neutral. – Ninad Munshi Jun 01 '23 at 18:12
  • "But now I can't integrate the first term", the OP asks for compute the first term ONLY, to compute the first part, it must involves Catalan's constant. Why do you say it is un-relevant to OP? @NinadMunshi – MathFail Jun 01 '23 at 18:14
  • I too didn't downvote – SavageTomato Jun 01 '23 at 18:16
  • The history shows 2 downvotes, strange. – MathFail Jun 01 '23 at 18:17
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    Context is important pedagogically. There are lots of times people answer a direct question from OP that I don't agree with. In reality many OPs don't realize they've gone down the wrong rabbit hole and that the OPs don't realize that they don't really want the answer to the direct question they asked. Back when you were a calculus 1 student, would you have appreciated an answer to a question you were told to do, that appealed to a special constant you had never seen before? I certainly would have not, and felt deflated that the answer was a trick outside my toolbox. – Ninad Munshi Jun 01 '23 at 18:18
  • How do you know this is a calculus 1 problem? The auther never mentions what level for this question. Learning is to learn something new, something unknown, something outside own's toolbox, isn't it? @NinadMunshi – MathFail Jun 01 '23 at 18:24
  • @MathFail, NinadMunshi is true that it is a calculus 1 problem and that i am new to definite integration and it properties. But u r also true to say that we need to know something new outside our toolbox. – SavageTomato Jun 01 '23 at 18:27
  • But when you post this OP, you didn't mention any background levels for this problem. You only ask for how to integrate the first part, so I answer you with the solution of the first part. @SavageTomato – MathFail Jun 01 '23 at 18:30
  • @MathFail sorry for that. I will take care of it next time. – SavageTomato Jun 01 '23 at 18:35
1

No need to evaluate the second term of OP’s substitution $$ \begin{aligned} \int_0^{\frac{\pi}{4}} \theta^2 \sec ^2 \theta d \theta=&\int_0^{\frac{\pi}{4}} \theta^2 d(\tan \theta) \\ = & {\left[\theta^2 \tan \theta\right]_0^{\frac{\pi}{4}}-2 \int_0^{\frac{\pi}{4}} \theta \tan \theta d \theta } \\ = & \frac{\pi^2}{16}+2 \int_0^{\frac{\pi}{4}} \theta d \ln (\cos \theta) \\ = & \frac{\pi^2}{16}+2[\theta \ln \cos \theta]_0^{\frac{\pi}{4}}+2 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta \\ = & \frac{\pi^2}{16}-\frac{\pi}{4} \ln 2+2J \end{aligned} $$ $$ \begin{aligned} I & =\int_0^{\frac{\pi}{4}} \theta^2 \sec ^2 \theta d \theta-2 \mathrm{~J} \\ & =\frac{\pi^2}{16}-\frac{\pi}{4} \ln 2 \end{aligned} $$

Lai
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0

$$ \int_0^1\left[\left(\tan ^{-1} x\right)^2-\frac{\ln \left(1+x^2\right)}{1+x^2}\right] d x $$

$$ \int_0^1\left(\tan ^{-1} x\right)^2 dx -\int_0^1\frac{\ln \left(1+x^2\right)}{1+x^2}d x $$

$$x\left(\tan ^{-1} x\right)^2- 2\int_0^1\frac{x(\tan^{-1}x)}{1+x^2}d x $$

Repeat this for the second integral

Kamal Saleh
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NadiKeUssPar
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