Question

1. integration cosecx .dx =log | tan (x/2) |+ c ​

Answer 1

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

Prove that

$\rm :\longmapsto\:\displaystyle\int\tt cosecx \: dx \: = \: log |tan\dfrac{x}{2} | + c$

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\tt cosecx \: dx$

On multiply cosecx + cotx, we get

$\rm \: = \: \: \displaystyle\int\tt cosecx \times \dfrac{cosecx + cotx}{cosecx + cotx} \: dx$

$\rm \: = \: \: \displaystyle\int\tt \dfrac{ {cosec}^{2} x + cosecx \: cotx}{cosecx + cotx} \: dx$

$\red{\rm :\longmapsto\:cosecx + cotx = y}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}(cosecx + cotx) = \dfrac{d}{dx}y$

$\rm :\longmapsto\: - cosecxcotx - {cot}^{2}x = \dfrac{dy}{dx}$

$\rm :\longmapsto\: - (cosecxcotx + {cosec}^{2}x)dx = dy$

$\rm :\longmapsto\: (cosecxcotx + {cosec}^{2}x)dx \: = \: - \: dy$

So, now given integral reduced to

$\rm \: = \: \: - \: \displaystyle\int\tt \frac{dy}{y}$

$\rm \: = \: \: - log |y| + c$

$\rm \: = \: \: log | {y}^{ - 1} | + c$

$\rm \: = \: \: log | \dfrac{1}{y} | + c$

$\rm \: = \: \: log \bigg | \dfrac{1}{y} \bigg| + c$

$\rm \: = \: \: log \bigg | \dfrac{1}{cosecx + cotx} \bigg| + c$

$\rm \: = \: \: log \bigg | \dfrac{1}{\dfrac{1}{sinx} + \dfrac{cosx}{sinx} } \bigg| + c$

$\rm \: = \: \: log \bigg | \dfrac{sinx}{ 1 + cosx} \bigg| + c$

$\rm \: = \: \: log \bigg | \dfrac{2sin\dfrac{x}{2} cos\dfrac{x}{2} }{ {2cos}^{2} \dfrac{x}{2} } \bigg| + c$

$\rm \: = \: \: log \bigg | \dfrac{sin\dfrac{x}{2} }{cos\dfrac{x}{2} } \bigg| + c$

$\rm \: = \: \: log \bigg | tan\dfrac{x}{2} \bigg| + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\tt cosecx \: dx \: = \: log |tan\dfrac{x}{2} | + c$

Formula Used :-

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - cosecxcotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = - \: {cosec}^{2}x}}$

$\boxed{ \rm{ \displaystyle\int\tt \frac{1}{x}dx = log |x| + c}}$

$\boxed{ \rm{ sin2x = 2sinx \: cosx}}$

$\boxed{ \rm{ 1 + cos2x = {2cos}^{2}x}}$