Question

o to pie/2 log cot x dx


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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm log(cotx) \: dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^{ \dfrac{\pi}{2} }\rm log(cotx) \: dx - - - (1)$

We know,

$\boxed{ \sf \: \displaystyle\int_0^{a}\rm \: f(x) \: dx \: = \: \displaystyle\int_0^{a}\rm f(a - x) \: dx}$

So, above integral can be rewritten as

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^{ \dfrac{\pi}{2} }\rm logcot \bigg[ \frac{\pi}{2} - x\bigg] \: dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^{ \dfrac{\pi}{2} }\rm log \: tanx \: dx - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2I \: =\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm log \: cotx \: dx + \displaystyle\int_0^{ \dfrac{\pi}{2} }\rm log \: tanx \: dx$

$\rm :\longmapsto\:2I \: =\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm \bigg[ log \: cotx + log \: tanx\bigg]\: dx$

We know,

$\boxed{ \sf \: logx \: + \: logy \: = \: logxy}$

So, using this, we get

$\rm :\longmapsto\:2I \: =\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm \bigg[ log \:( cotx \times tanx)\bigg]\: dx$

We know,

$\boxed{ \sf \: tanx = \frac{1}{cotx}} \: \rm :\implies\:\boxed{ \sf \: tanx \: \times \: cotx \: = \: 1}$

$\rm :\longmapsto\:2I \: =\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm \bigg[ log \:( 1)\bigg]\: dx$

We know,

$\boxed{ \bf \: log1 \: = \: 0}$

So, using this we get,

$\rm :\longmapsto\:2I \: =\displaystyle\int_0^{ \dfrac{\pi}{2} }\rm \bigg[ 0\bigg]\: dx$

$\rm :\longmapsto\:2I \: =0$

$\rm :\longmapsto\:I \: =0$

Hence,

$\bf\implies \:\:\displaystyle\int_0^{ \dfrac{\pi}{2} }\bf log(cotx) \: dx = 0$

Additional Information :-

$\boxed{ \sf \: \displaystyle\int_a^{b}\rm f(x) \: dx \: = \: \displaystyle\int_a^{b}\rm f(y) \: dy}$

$\boxed{ \sf \: \displaystyle\int_a^{b}\rm f(x) \: dx \: = \: - \: \displaystyle\int_b^{a}\rm f(x) \: dx}$

$\boxed{ \sf \: \displaystyle\int_a^{b}\rm f(x) \: dx = \displaystyle\int_a^{b}\rm f(a + b - x) \: dx}$

$\boxed{ \sf \: \displaystyle\int_0^{ 2a }\rm log \: f(x) \: dx = 0, \: \: if \: f(2a - x) = - f(x)}$

$\boxed{ \sf \: \displaystyle\int_0^{ 2a }\rm log \: f(x) \: dx =2 \displaystyle\int_0^{a}\rm f(x) \: dx, \:if \: f(2a - x) = f(x)}$