Question

Show by differentiating (and then using algebra) that $-\cot \left({\sin }^{-1}x\right)$and $-\frac{\sqrt{1-x^2}}{x}$are both antiderivatives of $\frac{1}{x^2\sqrt{1-x^2}}$. How can these two very different-looking functions be an antiderivative of the same function?

Short answer

Ans: By differentiating (and then using algebra) $-\cot \left({\sin }^{-1}x\right)\text{and}-\frac{\sqrt{1-x^2}}{x}$ both are antiderivatives of $\frac{1}{x^2\sqrt{1-x^2}}$

Step-by-step solution

Step 1. Given information.

given expresiion,

$-\cot \left({\sin }^{-1}x\right)\text{and}-\frac{\sqrt{1-x^2}}{x}$

Step 2. The objective is to show that −cot⁡sin−1⁡x is antiderivatives of 1x21−x2

The differentiation is,

$\begin{array}{r}-\cot \left({\sin }^{-1}x\right)\\ =\frac{d}{dx}\left[-\cot \left({\sin }^{-1}x\right)\right]\\ =-\cot \left(\frac{d}{dx}\left({\sin }^{-1}x\right)\right)\\ =-\cot \left(\frac{1}{\sqrt{1-x^2}}\right)\\ =\frac{1}{x^2\sqrt{1-x^2}}\end{array}$

Hence the expression is proved.

Step 3. The objective is to show that −cot⁡sin−1⁡x is antiderivatives of 1x21−x2

The differentiation is,

$\begin{array}{r}-\frac{\sqrt{1-x^2}}{x}\\ ={\left(-\frac{\sqrt{1-x^2}}{x}\right)}^{\mathrm{\prime }}\\ =-{\left(\frac{\sqrt{1-x^2}}{x}\right)}^{\mathrm{\prime }}\\ =-\frac{1}{x^2}\left[{\left(\sqrt{-x^2+1}\right)}^{\mathrm{\prime }}x-\sqrt{-x^2+1}{x}^{\mathrm{\prime }}\right]\\ =\frac{1}{x^2\sqrt{1-x^2}}\end{array}$

Hence the expression is proved.