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Evaluate: $$\int{\cos{\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)}dx}.$$

  • Method 1 Put $x=\cos(2\theta)$; then $\mathrm dx=-2\sin(2\theta)\,\mathrm d\theta$ $$\begin{align*} I&= \int\cos(2\cot^{-1}\tan\theta)(-2\sin(2\theta))\,\mathrm d\theta\\ &= \int {\cos{\left(2{\left(\frac{π}2-\tan^{-1}(\tan\theta)\right)}\right)}(-2\sin(2\theta))\,\mathrm d\theta}\\ &= \int {-2\sin(2\theta)\cos(π-2\theta)\,\mathrm d\theta} =\int{\sin(4\theta)\,\mathrm d\theta}\\ &=-\frac{1}{4}(2\cos^2(2\theta)-1)+c_1\\ &=- \frac{1}{2} x^2 + c\end{align*}$$

  • Method 2 $$\begin{align*} I &= \int{\cos{\left(2{\left(\frac{π}{2}-\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right)}\right)}\,\mathrm dx}\\ & = \int {-\cos{\left(2\tan^{-1}\sqrt{\frac{1-x}{x+1}}\right)}\,\mathrm dx}\\ &= \int {-\cos{\left(\tan^{-1}{\frac{\sqrt{1-x^2}}{x}}\right)}\,\mathrm dx}\\ &= \int -{\cos(\cos^{-1}(x))\,\mathrm dx}\\ &=- \frac{1}{2} x^2 + c\end{align*}$$

  • Method 3 As $$\cot^{-1}x= \begin{cases}\tan^{-1}\frac{1}{x}, & x>0\\π+\tan^{-1}\frac{1}{x}, &x<0 \end{cases} $$ taking case $1$ or $2$ : $$\begin{align*}I&= \int{\cos{\left(2\tan^{-1}\sqrt{\frac{1+x}{1-x}}\right)}\,\mathrm dx}\\ & = \int{\cos{\left(-\tan^{-1}{\frac{\sqrt{1-x^{2}}}{x}}\right)}\,\mathrm dx}\\ &= \int {\cos(\cos^{-1}(x))\,\mathrm dx}\\ &= \frac{1}{2} x^2 + c\end{align*}$$

Why does the answer of Method 3 not have negative sign? (I know that the answer of Method 1 and 2 is correct.)

ultralegend5385
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Daksh
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1 Answers1

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You made two mistakes in Method 3:

  • In case 1 ($x>0$), $2\tan^{-1}\sqrt{\frac{1+x}{1-x}}\in(0,\pi)$ is not equal to $-\tan^{-1}\frac{\sqrt{1-x^2}}x\in(-\frac\pi2,0)$, but to $-\tan^{-1}\frac{\sqrt{1-x^2}}x+\pi$.
  • In case 2 ($x<0$), $-\tan^{-1}\frac{\sqrt{1-x^2}}x\in(0,\frac\pi2)$ is not equal to $\cos^{-1}(x)\in(\frac\pi2,\pi)$, but to $\cos^{-1}(-x)$.
Anne Bauval
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