NIMCET 2026 — Mathematics PYQ

To solve the integral ${\int }_{-1}^1{f}^{-1}(y)dy$ as shown in image_c6093d.png, we first need to find the expression for the inverse function $f^{-1}(y)$.

Step 1: Find the inverse function $f^{-1}(y)$

Given $y=f(x)=\frac{x^2-1}{x^2+1}$, we solve for $x$ in terms of $y$:

$$y(x^2+1)=x^2-1$$

$$y{x}^2+y=x^2-1$$

$$1+y=x^2-y{x}^2$$

$$1+y=x^2(1-y)$$

$$x^2=\frac{1+y}{1-y}$$

Since the domain of $f$ is $[0,\infty )$, we take the positive square root:

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

Step 2: Set up the integral

We need to evaluate $I={\int }_{-1}^1\sqrt{\frac{1+y}{1-y}}dy$.

To solve this, we use the trigonometric substitution $y=\cos \theta$, which implies $dy=-\sin \theta d\theta$.

• When $y=-1$, $\theta =\pi$.

• When $y=1$, $\theta =0$.

Substituting these into the integral:

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

Step 3: Simplify and Evaluate

Using trigonometric identities $\frac{1+\cos \theta }{1-\cos \theta }=\frac{2{\cos }^2(\theta /2)}{2{\sin }^2(\theta /2)}={\cot }^2(\theta /2)$ and $\sin \theta =2\sin (\theta /2)\cos (\theta /2)$:

$$I={\int }_0^{\pi }\cot (\theta /2)\cdot (2\sin (\theta /2)\cos (\theta /2))d\theta$$

$$I={\int }_0^{\pi }\left(\frac{\cos (\theta /2)}{\sin (\theta /2)}\right)\cdot 2\sin (\theta /2)\cos (\theta /2)d\theta$$

$$I={\int }_0^{\pi }2{\cos }^2(\theta /2)d\theta$$

Using the identity $2{\cos }^2(\theta /2)=1+\cos \theta$:

$$I={\int }_0^{\pi }(1+\cos \theta )d\theta$$

$$I=[\theta +\sin \theta {]}_0^{\pi }$$

$$I=(\pi +\sin \pi )-(0+\sin 0)$$

$$I=\pi +0-0=\pi$$

Therefore, the value of the integral is $\pi$.