Question:
If \(f:[0,\infty)\to\mathbb R\),
\[
f(x)=\frac{x^2-1}{x^2+1},
\]
then find
\[
\int_{-1}^{1}f^{-1}(y)\,dy.
\]
If \(f:[0,\infty)\to\mathbb R\),
\[
f(x)=\frac{x^2-1}{x^2+1},
\]
then find
\[
\int_{-1}^{1}f^{-1}(y)\,dy.
\]
Show Hint
Whenever
\[
\frac{1+y}{1-y}
\]
appears under a square root, the substitution
\[
y=\cos 2\theta
\]
usually simplifies the integral dramatically.
Updated On: Jun 11, 2026
- \(\pi\)
- \(2\pi\)
- \(0\)
- \(\frac{\pi}{2}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the inverse function.
Let \[ y=\frac{x^2-1}{x^2+1}. \] Then \[ yx^2+y=x^2-1. \] Hence \[ x^2(1-y)=1+y. \] Therefore \[ x^2=\frac{1+y}{1-y}. \] Since \(x\ge0\), \[ f^{-1}(y) = \sqrt{\frac{1+y}{1-y}}. \]
Step 2: Evaluate the integral.
Put \[ y=\cos 2\theta. \] Then \[ \sqrt{\frac{1+y}{1-y}} = \cot\theta. \] Also \[ dy=-4\sin\theta\cos\theta\,d\theta. \] Thus \[ I = \int_{-1}^{1} \sqrt{\frac{1+y}{1-y}}\,dy = \frac{\pi}{2}. \] Therefore, \[ \boxed{\frac{\pi}{2}}. \]
Let \[ y=\frac{x^2-1}{x^2+1}. \] Then \[ yx^2+y=x^2-1. \] Hence \[ x^2(1-y)=1+y. \] Therefore \[ x^2=\frac{1+y}{1-y}. \] Since \(x\ge0\), \[ f^{-1}(y) = \sqrt{\frac{1+y}{1-y}}. \]
Step 2: Evaluate the integral.
Put \[ y=\cos 2\theta. \] Then \[ \sqrt{\frac{1+y}{1-y}} = \cot\theta. \] Also \[ dy=-4\sin\theta\cos\theta\,d\theta. \] Thus \[ I = \int_{-1}^{1} \sqrt{\frac{1+y}{1-y}}\,dy = \frac{\pi}{2}. \] Therefore, \[ \boxed{\frac{\pi}{2}}. \]
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