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Evaluate

$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$

This integral came up while solving another problem on Integration. Unfortunately, I'm quite bad at Inverse Trigonometry and was unable to proceed with it. The only thought which came to my mind was replacing $x$ with $\sin(\theta)$. Could somebody please help integrate this without using Differentiation under the Integral Sign? Many thanks in advance! $$$$EDIT: For using Differentiating under the Integral Sign, the suggested method was as follows: $$I(m)=\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$ Wethen differentiate with respect to $m$. Next, substitute $x=\sin (u)$ and then multiply the numerator and denominator by $\sec(u)$. The rest should be straightforward.

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    $\begingroup$ Wolfram Alpha seems to want to integrate by parts. $\endgroup$ – user7530 Jun 9 '15 at 7:04
  • $\begingroup$ I tried that way taking it as $$\int_{-1}^1 x'\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx$$ Unfortunately it did not work for me. $\endgroup$ – Ishan Jun 9 '15 at 7:07
  • $\begingroup$ Sir, I did it this way: $$x\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg )|_{-1}^{1} -\int_{-1}^1 x\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg )' dx$$ Taking $\int_{-1}^1 x\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg )' dx$ as J, $$J= x\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg )|_{-1}^{1}-\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx$$ Sir, this didn't seem to lead to anything I could work on. EDIT: $$\tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg )' = \dfrac{x}{\sqrt{1-x^2}(2-x^2)}$$ $\endgroup$ – Ishan Jun 9 '15 at 7:14
  • $\begingroup$ What happened to the derivative of the $\tan^{-1}$ term? Take the derivative and you will find it becomes more pleasant. $\endgroup$ – user7530 Jun 9 '15 at 7:16
  • $\begingroup$ @user7530 Sir I just edited my previous post. Please could you tell me what to do next, Sir? $\endgroup$ – Ishan Jun 9 '15 at 7:26
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First, integrate by parts to reduce the problem to calculating $$\int \frac{x^2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$ now split into two more manageable terms $$\int \frac{2}{\sqrt{1-x^2}(2-x^2)}\,dx + \int \frac{x^2-2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$ The left term is the only tricky one. Substitute $x= \sin u$ to get rid of the square root $$\int \frac{2}{2-\sin^2 u}\,du = \int \frac{2}{2\cos^2 u+\sin^2 u}\,du = \int \frac{2\sec^2 u}{2+\tan^2 u}\,du$$ and finally substitute $v = \frac{\tan u}{\sqrt{2}}$.

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  • $\begingroup$ Thanks very much Sir! Sir could you just explain the transition between $$2-\sin^2u=2\cos^2u+\sin^2u$$ $\endgroup$ – Ishan Jun 9 '15 at 8:10
  • $\begingroup$ @BetterWorld Use $1 = \cos^2u + \sin^2u$. $\endgroup$ – user7530 Jun 9 '15 at 8:12
  • $\begingroup$ Oh, I got it Sir. $$2=2(\sin^2u+\cos^2u)$$ $\endgroup$ – Ishan Jun 9 '15 at 8:12
  • $\begingroup$ We wrote it nearly at the same time:) Sir, I'm really grateful to you. Thanks very much for your help! $\endgroup$ – Ishan Jun 9 '15 at 8:13
  • $\begingroup$ Sir could I just ask one last question? Sir what motivated you to solve the integral the way you did? It's just that I'm trying to understand what methods to use for different types of Integrals. $\endgroup$ – Ishan Jun 9 '15 at 8:14
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$$\begin{eqnarray*}\int_{-1}^{1}\arctan\left(\frac{1}{\sqrt{1-x^2}}\right)\,dx &=& \pi-2\int_{0}^{1}\arctan(\sqrt{1-x^2})\\&=&\pi-2\int_{0}^{\pi/2}\cos(\theta)\arctan(\cos\theta)\,d\theta\\&=&\pi-2\int_{0}^{\pi/2}\frac{\sin(\theta)^2}{1+\cos^2\theta}\,d\theta\\&=&\pi-2\int_{0}^{+\infty}\frac{t^2\,dt}{(1+t^2)(2+t^2)}\\&=&\pi-2\int_{0}^{+\infty}\left(\frac{2}{t^2+2}-\frac{1}{t^2+1}\right)\,dt\\&=&\color{red}{(2-\sqrt{2})\pi}.\end{eqnarray*}$$ Steps involved:

  • $\arctan\frac{1}{t}=\frac{\pi}{2}-\arctan t$
  • substitution $x=\sin\theta$
  • integration by parts
  • substitution $\theta=\arctan t$
  • partial fraction decomposition
  • profit.
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  • $\begingroup$ @BetterWorld: you're welcome. $\endgroup$ – Jack D'Aurizio Jun 9 '15 at 8:39

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