Question

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

Answer 1

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

➢ Given integral is

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

can be further rewritten as

$\rm \: = \: \: \displaystyle\int\sf \dfrac{1}{ {cos}^{2}2x \: {(cotx - tanx)}^{2} } \: dx$

Now, Consider,

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

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

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

$\rm \: = \: \: 2 \times \dfrac{cos2x}{2sinxcosx}$

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

$\bf\implies \:cotx - tanx= \: \: \dfrac{2cos2x}{sin2x}$

So, given integral reduced to

$\rm \: = \: \: \displaystyle\int\sf \dfrac{1}{ {cos}^{2}2x \: {(cotx - tanx)}^{2} } \: dx$

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

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

Now, we use method of Substitution to evaluate such integral,

Substitute,

$\red{\rm :\longmapsto\:tan2x = y}$

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

$\red{\rm :\longmapsto\: {sec}^{2} 2x \times 2 \: dx = dy}$

$\red{\rm :\longmapsto\: {sec}^{2} 2x \: dx = \dfrac{1}{2} \: dy}$

On substituting these values, we get

$\rm \: = \: \: \dfrac{1}{8}\displaystyle\int\sf {y}^{2} \: dy$

$\rm \: = \: \: \dfrac{1}{8} \: \times \dfrac{ {y}^{2 + 1} }{2 + 1} + c$

$\rm \: = \: \: \dfrac{1}{8} \: \times \dfrac{ {y}^{3} }{3} + c$

$\rm \: = \: \: \dfrac{ {y}^{3} }{24} + c$

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

Hence,

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

Formula Used :-

$\blue{ \boxed{\bf \:cos2x = {cos}^{2}x - {sin}^{2}x}}$

$\blue{ \boxed{\bf \:sin2x = 2sinx \: cosx}}$

$\blue{ \boxed{\bf \:\dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\blue{ \boxed{\bf \:\dfrac{d}{dx}x =1}}$

$\blue{ \boxed{\bf \:\displaystyle\int\bf \: {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

Additional Information :-

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: cosxdx = sinx + c}}$

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: sinxdx = - \: cosx + c}}$

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: cosecx \: cotx \: dx = - \: cosecx + c}}$

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: secx \: tanx \: dx = \: secx + c}}$

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: {sec}^{2} xdx = \: tanx + c}}$

$\blue{ \boxed{\bf \:\displaystyle\int\sf \: {cosec}^{2} xdx = - \: cotx + c}}$