4

Evaluate

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

Could somebody please help integrate this without using Differentiation under the Integral Sign?

User1234
  • 3,958

3 Answers3

12

$$\begin{eqnarray*}\int_{-1}^{1}\arctan\left(\frac{1}{\sqrt{1-x^2}}\right)\,dx &=& \pi-2\int_{0}^{1}\arctan(\sqrt{1-x^2})\\&=&\pi-2\int_{0}^{\pi/2}\cos(\theta)\arctan(\cos\theta)\,d\theta\\&=&\pi-2\int_{0}^{\pi/2}\frac{\sin(\theta)^2}{1+\cos^2\theta}\,d\theta\\&=&\pi-2\int_{0}^{+\infty}\frac{t^2\,dt}{(1+t^2)(2+t^2)}\\&=&\pi-2\int_{0}^{+\infty}\left(\frac{2}{t^2+2}-\frac{1}{t^2+1}\right)\,dt\\&=&\color{red}{(2-\sqrt{2})\pi}.\end{eqnarray*}$$ 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.
Jack D'Aurizio
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5

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 $v = \frac{\tan u}{\sqrt{2}}$.

user7530
  • 49,280
  • @BetterWorld Use $1 = \cos^2u + \sin^2u$. – user7530 Jun 09 '15 at 08:12
  • Everything up to the last step was motivated by simplifying the integral -- removing inverse trig functions, removing square roots, and decreasing the powers of the terms involved. The last step -- the $v$ substitution -- I admit was ad hoc. I stared at $\frac{1}{2-\sin^2 u}$ and tried various manipulations until something worked. – user7530 Jun 09 '15 at 08:30
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$$ \begin{aligned} I &=\int_{-1}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\ &=2 \int_{0}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\ & \stackrel{IBP}{=} 2\left[\tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)\right]_{0}^{1}-2 \int_{0}^{1} x \cdot\frac{1}{1+\left(\frac{1}{\sqrt{1-x^{2}}}\right)^{2}}\left(\frac{x}{\left(-x^{2}\right)^{\frac{3}{2}}}d x \right)\\ &=\pi-2 \underbrace{\int_{0}^{1} \frac{x^{2}}{\left(2-x^{2}\right) \sqrt{1-x^{2}}} d x}_{J} \end{aligned} $$ Let $x=\sin \theta$, where $0 \leqslant \theta \leqslant \dfrac{\pi}{2},$ then $$ \begin{aligned} J &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{\left(2-\sin ^{2} \theta\right) \cos \theta} \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{2-\sin ^{2} \theta} d \theta \\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{2}{2-\sin ^{2} \theta}-1\right) d \theta \\ &=2 \int_{0}^{\frac{\pi}{2}} \frac{\csc ^{2} \theta}{2 \csc ^{2} \theta-1} d \theta-\frac{\pi}{2} \\ &=-2 \int_{0}^{\frac{\pi}{2}} \frac{d(\cot \theta)}{1+2 \cot ^{2} \theta}-\frac{\pi}{2}\\ &=-2 \cdot \frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} \cot \theta)\right]_{0}^{\frac{\pi}{2}}-\frac{\pi}{2} \\ &=\frac{\pi \sqrt{2}}{2}-\frac{\pi}{2} \end{aligned} $$ Now we can conclude that $I=\pi(2-\sqrt{2})$.

Lai
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