NIMCET 2025 — Mathematics PYQ

1. Step 1 Rewrite the integrand. 

• The integrand is $\frac{1+2\cos x}{(2+\cos x{)}^2}$. 

• Rewrite the numerator: $1+2\cos x=(2+\cos x)-1+\cos x=(2+\cos x)+(-\sin x)\cdot \frac{1}{-\sin x}+\cos x$. 

• This form is not directly helpful for a simple substitution. 

 

2. Step 2 Consider the derivative of $\frac{1}{2+\cos x}$. 

• Let $f(x)=2+\cos x$. 

• Then $f^{\prime }(x)=-\sin x$. 

• The derivative of $\frac{1}{2+\cos x}$ is $-\frac{-\sin x}{(2+\cos x{)}^2}=\frac{\sin x}{(2+\cos x{)}^2}$. 

 

3. Step 3 Manipulate the integrand to match a known derivative. 

• We want to integrate $\frac{1+2\cos x}{(2+\cos x{)}^2}$. 

• Consider the derivative of $\frac{\sin x}{2+\cos x}$. 

• Using the quotient rule, $\frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right)=\frac{\cos x(2+\cos x)-\sin x(-\sin x)}{(2+\cos x{)}^2}$. 

• This simplifies to $\frac{2\cos x+{\cos }^{2}x+{\sin }^{2}x}{(2+\cos x{)}^2}=\frac{2\cos x+1}{(2+\cos x{)}^2}$. 

• This is exactly the integrand. 

 

4. Step 4 Evaluate the definite integral. 

• ${\int }_0^{\frac{\pi }{2}}\frac{1+2\cos x}{(2+\cos x{)}^2}dx={\left[\frac{\sin x}{2+\cos x}\right]}_0^{\frac{\pi }{2}}$. 

• Evaluate at the upper limit: $\frac{\sin (\frac{\pi }{2})}{2+\cos (\frac{\pi }{2})}=\frac{1}{2+0}=\frac{1}{2}$. 

• Evaluate at the lower limit: $\frac{\sin (0)}{2+\cos (0)}=\frac{0}{2+1}=0$. 

• The value of the integral is $\frac{1}{2}-0=\frac{1}{2}$. [2]  

 

5. Step 5 Determine the interval. 

• The value of the integral is $\frac{1}{2}=0.5$. [2]  

• This value lies in the interval $(0,1)$. 

 

Solution 

The value of the integral is $\frac{1}{2}$, which lies in the interval $(0,1)$.