Aspire Faculty ID #18686 · Topic: NIMCET 2026 · Just now
NIMCET 2026

If $f:[0,\infty)\to R$ is defined by $f(x)=\frac{x^2-1}{x^2+1}$, find the value of: $\int_{-1}^{1} f^{-1}(y)dy$

Solution

Let 
$y=\frac{x^2-1}{x^2+1}$ 
$y(x^2+1)=x^2-1$ 
$yx^2+y=x^2-1$ 
$x^2(y-1)=-(1+y)$ 
$x^2=\frac{1+y}{1-y}$ 

Since $x\ge 0$, 
$f^{-1}(y)=\sqrt{\frac{1+y}{1-y}}$ 

Now, $I=\int_{-1}^{1}\sqrt{\frac{1+y}{1-y}}dy$ 
Put 
$y=\cos\theta$ 
$dy=-\sin\theta,d\theta$ 

When $y=-1,\ \theta=\pi$ 
When $y=1,\ \theta=0$ 

$I=\int_{\pi}^{0}\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}(-\sin\theta)d\theta$ 

$=\int_0^\pi \cot\frac{\theta}{2}\sin\theta d\theta$ 

Now, $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ 
$I=\int_0^\pi 2\cos^2\frac{\theta}{2} d\theta$ 
$=\int_0^\pi (1+\cos\theta) d\theta$ 
$=[\theta+\sin\theta]_0^\pi$ 
$=\pi$ 
Answer: $\pi$
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