Question

Please slove this problem.​

Answer 1

Given :

$\rm \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} = \dfrac{\pi}{2}$

To prove :

R.H.S = L.H.S

Proof :

$\rm \longrightarrow \: I = \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x}$

We know that :-

$\begin{gathered}\sf \displaystyle \int_{ - a}^{a} f(x) dx= \begin{cases} \sf{2 \displaystyle\int_{ 0}^{a} f(x) dx\: \: ;\: \rm if \: f(x) = even \: function } \\\ \\ \sf{0 \: \: \: ;\: \rm if \: f(x) = odd \: function} \end{cases}\end{gathered}$

$\rm \longrightarrow \: I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} \: \: \: \: [ let \: this \: be \: equation \: (1) ]$

We know that :-

$\boxed{\rm \bf \displaystyle\int ^{ a} _{0} \bf f(x) \: dx= \bf \displaystyle\int ^{ a} _{0} \bf f(a - x) \: dx}$

$\rm \longrightarrow \: I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4}( \frac{\pi}{2} - x) }$

$\rm \longrightarrow \: I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {tan}^{4}( x) } \\ \\ \rm \longrightarrow \: I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + \dfrac{1}{ {cot}^{4}( x) } }$

$\rm \longrightarrow \: I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{{cot}^{4}( x)}{1 + {cot}^{4}( x) } \: dx \: \: \: \: \: [ let \: this \: be \: equation \: (2) ]$

Now , by adding equation (1) and (2) we get :-

$\rm \longrightarrow \: 2I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} + \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{{cot}^{4}( x)}{1 + {cot}^{4}( x) } \: dx$

$\rm \longrightarrow \: 2I = \displaystyle2 \int_{ 0}^{ \frac{\pi}{2} } \rm\frac{1 + {cot}^{4} x}{1 + {cot}^{4} x} \: dx$

$\rm \longrightarrow \: I = \displaystyle\int_{ 0}^{ \frac{\pi}{2} } 1 \rm \: dx$

$\rm \longrightarrow \: I = \displaystyle \bigg[ \: \rm \: x \: \: \bigg]_{ 0}^{ \frac{\pi}{2} }$

$\rm \longrightarrow \: I = ( \dfrac{\pi}{2} - 0)$

$\rm \longrightarrow \: I = \dfrac{\pi}{2}$

$\therefore \rm \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} = \dfrac{\pi}{2}$

Hence proved

Answer 2

$\large\underline\blue{\bold{Given \:Question - }}$

$\bull \: \: \: \rm \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} = \: \dfrac{\pi}{2}$

$\begin{gathered}\Large{\bold{ \sf \: \green{\underline{Formula \: Used \::}}}} \end{gathered}$

$\red{\begin{gathered}\begin{gathered}\sf \displaystyle \int_{ - a}^{a} f(x) dx= \begin{cases} \rm{2 \displaystyle\int_{ 0}^{a} f(x) dx\: \: ;\: \rm if \: f(x) = even \: function } \\\ \\ \sf{0 \: \: \: ;\: \rm if \: f(x) = odd \: function} \end{cases}\end{gathered}\end{gathered}}$

$\pink{\rm \bf \displaystyle\int ^{ a} _{0} \bf f(x) \: dx= \bf \displaystyle\int ^{ a} _{0} \bf f(a - x) \: dx}$

$\large\underline\purple{\bold{Solution :- }}$

$\bf \: Let \: I \: = \: \rm \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} - - (1)$

Now,

$\longmapsto \red{ \bf \: Let \: f(x) = \dfrac{1}{1 + {cot}^{4}x } }$

So,

$\longmapsto { \bf \: \: f( - x) = \dfrac{1}{1 + {cot}^{4}( - x) } }$

$\longmapsto { \bf \: \: f( - x) = \dfrac{1}{1 + {cot}^{4}x } }$

$\longmapsto { \bf \:\: f( - x) = f(x)}$

$\rm :\implies\:\: \boxed{ \pink{\bf \: f(x) \: is \: even \: function}}$

Hence,

$\rm :\implies\:\rm \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} = \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x}$

$\rm :\implies\:I \: = \: \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} x} - - - (2)$

$\longmapsto\: \boxed{ \pink{\bf \:change \: x \: \to \: \dfrac{\pi}{2} - x }}$

$\rm :\implies\:I \: = \: \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {cot}^{4} \bigg(\dfrac{\pi}{2} - x \bigg)}$

$\rm :\implies\:I \: = \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{dx}{1 + {tan}^{4} x}$

$\rm :\implies\:I \: = \: \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{ {cot}^{4}x \: dx}{1 + {cot}^{4} x} - - (3)$

Now,

on adding equation (2) and equation (3), we get

$\rm :\implies\:2I \: = \: \rm \displaystyle 2\int_{ 0}^{ \frac{\pi}{2} } \rm\frac{(1 + {cot}^{4} x )\: dx}{1 + {cot}^{4} x}$

$\rm :\implies\:I \: = \: \rm \displaystyle \int_{ 0}^{ \frac{\pi}{2} } 1 \: dx$

$\rm :\implies\:I \: = \bigg[ \: x \: \bigg]_0^{ \frac{\pi}{2} }$

$\bf\implies \:I \: = \: \dfrac{\pi}{2}$

Hence,

$\rm :\implies\: \boxed{ \green{\bf \displaystyle \int_{ - \frac{\pi}{2}}^{ \frac{\pi}{2} } \bf\frac{dx}{1 + {cot}^{4} x} = \dfrac{\pi}{2} }}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$