If $$\int_0^{\pi/2} \frac{\cot x}{\cot x + \text{cosec} x} dx = m(\pi + n)$$, then mn is equal to

We have the integral

$$I \;=\;\int_{0}^{\pi/2}\dfrac{\cot x}{\cot x+\csc x}\,dx.$$

First we recall the trigonometric identities

$$\cot x=\dfrac{\cos x}{\sin x}\quad\text{and}\quad\csc x=\dfrac{1}{\sin x}.$$

Substituting these into the integrand gives

$$\cot x+\csc x=\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x} =\dfrac{\cos x+1}{\sin x}.$$

Hence the fraction inside the integral simplifies as

$$\dfrac{\cot x}{\cot x+\csc x} =\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x+1}{\sin x}} =\dfrac{\cos x}{\cos x+1}.$$

So the integral becomes

$$I=\int_{0}^{\pi/2}\dfrac{\cos x}{1+\cos x}\,dx.$$

To integrate this, we write the numerator in a convenient way:

$$\dfrac{\cos x}{1+\cos x} =\dfrac{(1+\cos x)-1}{1+\cos x} =1-\dfrac{1}{1+\cos x}.$$

Therefore

$$I=\int_{0}^{\pi/2}\Bigl[1-\dfrac{1}{1+\cos x}\Bigr]dx =\int_{0}^{\pi/2}1\,dx-\int_{0}^{\pi/2}\dfrac{dx}{1+\cos x}.$$

The first integral is immediate:

$$\int_{0}^{\pi/2}1\,dx=\dfrac{\pi}{2}.$$

For the second integral we use the identity $$1+\cos x=2\cos^2\!\bigl(\tfrac{x}{2}\bigr).$$ Thus

$$\int_{0}^{\pi/2}\dfrac{dx}{1+\cos x} =\int_{0}^{\pi/2}\dfrac{dx}{2\cos^2\!\bigl(\tfrac{x}{2}\bigr)} =\dfrac12\int_{0}^{\pi/2}\sec^2\!\bigl(\tfrac{x}{2}\bigr)dx.$$

Now we put $$t=\dfrac{x}{2}\;\Longrightarrow\;dx=2\,dt.$$ When $$x=0,\;t=0$$ and when $$x=\dfrac{\pi}{2},\;t=\dfrac{\pi}{4}.$$ Hence

$$\dfrac12\int_{0}^{\pi/2}\sec^2\!\bigl(\tfrac{x}{2}\bigr)dx =\dfrac12\int_{0}^{\pi/4}\sec^2 t\,(2\,dt) =\int_{0}^{\pi/4}\sec^2 t\,dt =\Bigl[\tan t\Bigr]_{0}^{\pi/4} =\tan\!\Bigl(\dfrac{\pi}{4}\Bigr)-\tan 0 =1-0 =1.$$

Collecting these results, we have

$$I=\dfrac{\pi}{2}-1.$$

The question states that

$$I = m(\pi+n).$$

Comparing $$\dfrac{\pi}{2}-1$$ with $$m(\pi+n),$$ we equate the coefficients of $$\pi$$ first:

$$m\pi=\dfrac{\pi}{2}\;\Longrightarrow\;m=\dfrac12.$$

Substituting $$m=\dfrac12$$ back, we get

$$\dfrac12(\pi+n)=\dfrac{\pi}{2}-1 \;\;\Longrightarrow\;\; \dfrac{n}{2}=-1 \;\;\Longrightarrow\;\; n=-2.$$

Therefore

$$mn=\Bigl(\dfrac12\Bigr)(-2)=-1.$$

Hence, the correct answer is Option C.