If \displaystyle\int\limits_0^\pi/2 \frac\cot x\cot x + \csc x \, dx = m(\pi + n) , then m \cdot n is equal to

Definite Integration PYQ — JEE Main 2019 · 12 Apr · Shift 1 · Q26 with Solution

View written solutionFree

Correct answer: A

We need to evaluate $I=\int_0^{\pi/2} \frac{\cot x}{\cot x+\csc x}\,dx.$

Using $\cot x=\frac{\cos x}{\sin x}, \qquad \csc x=\frac{1}{\sin x},$ we get $\cot x+\csc x=\frac{\cos x+1}{\sin x}.$

So,

=\frac{\frac{\cos x}{\sin x}}{\frac{\cos x+1}{\sin x}} =\frac{\cos x}{1+\cos x}.$$ Hence, $$I=\int_0^{\pi/2} \frac{\cos x}{1+\cos x}\,dx.$$ 2. **Rewrite the integrand** Observe that $$\frac{\cos x}{1+\cos x}=1-\frac{1}{1+\cos x}.$$ Therefore, $$I=\int_0^{\pi/2}1\,dx-\int_0^{\pi/2}\frac{1}{1+\cos x}\,dx.$$ 3. **Use the identity $1+\cos x=2\cos^2\frac{x}{2}$** Then $$\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}=\frac{1}{2}\sec^2\frac{x}{2}.$$ So, $$\int_0^{\pi/2}\frac{1}{1+\cos x}\,dx =\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}\,dx.$$ Let $$u=\frac{x}{2}\implies dx=2\,du.$$ When $x=0, u=0$ and when $x=\pi/2, u=\pi/4$. Thus, $$\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}\,dx =\frac{1}{2}\int_0^{\pi/4} \sec^2 u\cdot 2\,du =\int_0^{\pi/4}\sec^2 u\,du =\tan u\Big|_0^{\pi/4}=1.$$ 4. **Compute the integral** Also, $$\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.$$ Hence, $$I=\frac{\pi}{2}-1.$$ 5. **Match with the given form** Given $$I=m(\pi+n).$$ We write $$\frac{\pi}{2}-1=\frac{1}{2}(\pi-2).$$ So, $$m=\frac{1}{2}, \qquad n=-2.$$ Therefore, $$m\cdot n=\frac{1}{2}\cdot(-2)=-1.$$ 6. **Check options** The correct option is: $$\boxed{\text{A: }-1}$$

View written solutionFree

Correct answer: A

We need to evaluate $I=\int_0^{\pi/2} \frac{\cot x}{\cot x+\csc x}\,dx.$

Using $\cot x=\frac{\cos x}{\sin x}, \qquad \csc x=\frac{1}{\sin x},$ we get $\cot x+\csc x=\frac{\cos x+1}{\sin x}.$

So,

=\frac{\frac{\cos x}{\sin x}}{\frac{\cos x+1}{\sin x}} =\frac{\cos x}{1+\cos x}.$$ Hence, $$I=\int_0^{\pi/2} \frac{\cos x}{1+\cos x}\,dx.$$ 2. **Rewrite the integrand** Observe that $$\frac{\cos x}{1+\cos x}=1-\frac{1}{1+\cos x}.$$ Therefore, $$I=\int_0^{\pi/2}1\,dx-\int_0^{\pi/2}\frac{1}{1+\cos x}\,dx.$$ 3. **Use the identity $1+\cos x=2\cos^2\frac{x}{2}$** Then $$\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}=\frac{1}{2}\sec^2\frac{x}{2}.$$ So, $$\int_0^{\pi/2}\frac{1}{1+\cos x}\,dx =\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}\,dx.$$ Let $$u=\frac{x}{2}\implies dx=2\,du.$$ When $x=0, u=0$ and when $x=\pi/2, u=\pi/4$. Thus, $$\frac{1}{2}\int_0^{\pi/2}\sec^2\frac{x}{2}\,dx =\frac{1}{2}\int_0^{\pi/4} \sec^2 u\cdot 2\,du =\int_0^{\pi/4}\sec^2 u\,du =\tan u\Big|_0^{\pi/4}=1.$$ 4. **Compute the integral** Also, $$\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.$$ Hence, $$I=\frac{\pi}{2}-1.$$ 5. **Match with the given form** Given $$I=m(\pi+n).$$ We write $$\frac{\pi}{2}-1=\frac{1}{2}(\pi-2).$$ So, $$m=\frac{1}{2}, \qquad n=-2.$$ Therefore, $$m\cdot n=\frac{1}{2}\cdot(-2)=-1.$$ 6. **Check options** The correct option is: $$\boxed{\text{A: }-1}$$