Question

find the value of integral of sec^2 2x/(cotx-tanx)^2 dx​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \rm \int \frac{ {sec}^{2} 2x}{ {(tanx - cotx)}^{2} } \: dx \:$

Consider,

$\rm :\longmapsto\:tanx - cotx$

$\rm \: = \: \: \dfrac{cosx}{sinx} - \dfrac{sinx}{cosx}$

$\rm \: = \: \: \dfrac{ {cos}^{2} x - {sin}^{2} x}{cosx \: sinx}$

Multiply and divide by 2, we get

$\rm \: = \: \: \dfrac{ 2({cos}^{2} x - {sin}^{2} x)}{2 \: cosx \: sinx}$

We know,

$\red{\boxed{ \rm \: cos2x = {cos}^{2}x - {sin}^{2}x}}$

and

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

So, using this, we get

$\rm \: = \: \: \dfrac{2cos2x}{sin2x}$

$\purple{\rm :\longmapsto\:cotx - tanx= \: \: \dfrac{2cos2x}{sin2x}}$

Therefore,

$\rm :\longmapsto\: {(cotx - tanx)}^{2} = \dfrac{4 {cos}^{2}2x}{ {sin}^{2}2x}$

Now, Consider

$\rm :\longmapsto\:\displaystyle \rm \int \frac{ {sec}^{2} 2x}{ {(tanx - cotx)}^{2} } \: dx \:$

$\rm \: = \: \: \displaystyle \rm \int \dfrac{1}{ {cos}^{2}2x \times \dfrac{ {4cos}^{2} 2x}{ {sin}^{2} 2x} }$

$\rm \: = \: \: \displaystyle \rm \int \frac{{sin}^{2}2x }{4 {cos}^{2}2x \times {cos}^{2}2x} \: dx$

$\rm \: = \: \: \dfrac{1}{4}\displaystyle \rm \int {tan}^{2}2x \: {sec}^{2}2x \: dx$

can be rewritten as

$\rm \: = \: \: \dfrac{1}{8}\displaystyle \rm \int {tan}^{2}2x \: (2{sec}^{2}2x) \: dx$

We know,

$\red{\boxed{ \rm \: \displaystyle \rm \int {(f(x))}^{n} \: f'(x) \: dx \: = \: \frac{ {(f(x))}^{n + 1} }{n + 1} \: + \: c}}$

Here,

We have

$\rm :\longmapsto\:\dfrac{d}{dx}tan2x = 2{sec}^{2}2x$

So, using this, we get

$\rm \: = \: \: \dfrac{1}{8} \dfrac{ {(tan2x)}^{2 + 1} }{2 + 1} + c$

$\rm \: = \: \: \dfrac{1}{8} \dfrac{ {(tan2x)}^{3} }{3} + c$

$\rm \: = \: \: \dfrac{1}{24} {tan}^{3}2x + c$

Hence,

$\bf :\longmapsto\:\displaystyle \rm \int \frac{ {sec}^{2} 2x}{ {(tanx - cotx)}^{2} } \: dx \: = \dfrac{1}{24} {tan}^{3}2x + c$

Additional Information :-

$\red{\boxed{ \rm \:\displaystyle \rm \int \frac{f'(x)}{f(x)}dx = log(f(x)) + c}}$

$\red{\boxed{ \rm \: \displaystyle \rm \int \frac{f'(x)}{ \sqrt{f(x)} }dx \: = \: 2 \sqrt{f(x)} + c}}$