JEE Main 2019 — Mathematics Calculus

Let $\displaystyle I=\int _{0}^{\frac{\pi }{2}}\frac{\mathrm{cot}x}{\mathrm{cot}x+\mathrm{cosec}x}dx$I=∫02π​​cotx+cosecxcotx​dx

$$\Rightarrow I=\int _{0}^{\frac{\pi }{2}}\frac{cosx}{cosx+1}dx$$⇒I=∫02π​​cosx+1cosx​dx

$$\Rightarrow I=\int _{0}^{\frac{\pi }{2}}\frac{cosx+1-1}{cosx+1}dx$$⇒I=∫02π​​cosx+1cosx+1−1​dx

$$\Rightarrow I=\int _{0}^{\frac{\pi }{2}}(1-\frac{1}{2{\mathrm{cos}}^{2}\frac{x}{2}})dx$$⇒I=∫02π​​(1−2cos22x​1​)dx

$$\Rightarrow I=\int _{0}^{\frac{\pi }{2}}(1-\frac{1}{2}se{c}^{2}\frac{x}{2})dx$$⇒I=∫02π​​(1−21​sec22x​)dx

$$\Rightarrow I={(x-\mathrm{tan}\frac{x}{2})}_{0}^{\frac{\pi }{2}}$$⇒I=(x−tan2x​)02π​​

$$\Rightarrow I=(\frac{\pi }{2}-1)-(0-0)$$⇒I=(2π​−1)−(0−0)

$\Rightarrow I=\frac{\pi }{2}-1=\frac{1}{2}(\pi -2)$⇒I=2π​−1=21​(π−2).

Now using given information $\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\mathrm{cot}x}{\mathrm{cot}x+\mathrm{cosec}x}dx=m(\pi +n),$∫02π​​cotx+cosecxcotx​dx=m(π+n), clearly $m=\frac{1}{2}$m=21​ and $n=-2$n=−2.

$$\Rightarrow mn=-1$$⇒mn=−1