Math, asked by CadburyDarling, 6 hours ago

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

Answers

Answered by mathdude500
4

\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}}

Similar questions