JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 4)
If $$\int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx$$ = m($$\pi $$ + n), then m.n is equal to
- 1
1
$$ - {1 \over 2}$$
$${1 \over 2}$$
Explanation
$$\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx} $$
$$ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} $$
$$ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)$$
mn = $${1 \over 2}x - 2 = - 1$$
$$ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} $$
$$ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)$$
mn = $${1 \over 2}x - 2 = - 1$$
Comments (0)
