Question

integration [ (cosx+sinx) (1-1/2 sin2x) sec^2 x cosec^2 x dx]

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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm (sinx + cosx){\bigg(1 - \dfrac{1}{2}sin2x \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

To solve this integral, the following identities used :-

$\boxed{ \sf{ \:sin2x = 2sinxcosx}}$

$\boxed{ \sf{ \:(x + y)( {x}^{2} + {y}^{2} - xy) = {x}^{3} + {y}^{3}}}$

$\boxed{ \sf{ \:secx = \frac{1}{cosx}}}$

$\boxed{ \sf{ \:cosecx = \frac{1}{sinx}}}$

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

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

Let's solve the problem now!!

$\rm :\longmapsto\:\displaystyle\int\rm (sinx + cosx){\bigg(1 - \dfrac{1}{2}sin2x \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\rm \: = \:\displaystyle\int\rm (sinx + cosx){\bigg(1 - \dfrac{1}{2}2sinxcosx \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\rm \: = \:\displaystyle\int\rm (sinx + cosx){\bigg(1 - sinxcosx \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\rm \: = \:\displaystyle\int\rm (sinx + cosx){\bigg( {sin}^{2}x+{cos}^{2}x- sinxcosx \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\rm \: = \:\displaystyle\int\rm {\bigg( {sin}^{3}x+{cos}^{3}x \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\rm \: = \:\displaystyle\int\rm {\bigg( {sin}^{3}x+{cos}^{3}x \bigg) } \frac{1}{ {cos}^{2}x} \frac{1}{ {sin}^{2}x } \: dx$

$\rm \: = \:\displaystyle\int\rm {\bigg( \frac{ {sin}^{3} x}{ {cos}^{2}x {sin}^{2}x} + \frac{ {cos}^{3}x}{ {cos}^{2}x {sin}^{2}x} \bigg) } \: dx$

$\rm \: = \:\displaystyle\int\rm {\bigg( \frac{ {sin}^{} x}{ {cos}^{2}x } + \frac{ {cos}^{}x}{{sin}^{2}x} \bigg) } \: dx$

$\rm \:  =  \:  \:\displaystyle\int\rm {\bigg(secx \: tanx \: + \: cosecx \: cotx\bigg) } \: dx$

$\rm \:  =  \:  \:secx \: - \: cosecx \: + \: c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\rm (sinx + cosx){\bigg(1 - \dfrac{1}{2}sin2x \bigg) } {sec}^{2}x {cosec}^{2}x \: dx$

$\bf \:  =  \:  \red{\:secx \: - \: cosecx} \: + \: c$

Therefore,

• Option (c) is correct.

Additional Information :-

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

$\boxed{ \sf{ \:\displaystyle\int\rm cosx \: dx = \: sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm sinx \: dx = - \: cosx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm tanx \: dx = - \:log cosx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm cotx \: dx = \:log sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm cosecx \: dx = \:log (cosecx - cotx) + c}}$

$\boxed{ \sf{ \:\displaystyle\int\rm secx \: dx = \:log (secx + tanx) + c}}$

Answer 2

$\large\underline{\red{\mathfrak{Solution\;:}}}$

$\longmapsto\displaystyle\int \sf (\cos x + \sin x )(1-\frac{1}{2}\sin2x )(\sec^2x cosec^2x )dx$

$\red \;\;\bigstar\;\;\boxed{\green{\sin2x= 2\sin x\cos x}}$

$\\\\\\\implies\sf\displaystyle\int \sf (\cos x + \sin x )(1-\frac{1}{\cancel{2}}\times\cancel{2}\sin x\cos x )(\sec^2xcosec^2x )dx\\\\\implies\displaystyle\int \sf (\cos x + \sin x )(1-\sin x\cos x )(\sec^2xcosec^2x )dx$

$\\\\\red\bigstar\;\;\boxed{\sin^2\theta+\cos^2\theta = 1}$

$\\\\\\\implies\displaystyle\int \sf (\cos x + \sin x )(\blue{\sin^2\theta+\cos^2\theta}-\sin x\cos x )(\sec^2xcosec^2x )dx$

$\\\red\bigstar\;\;\boxed{\sf\green{(a+b)(a^2+b^2-ab)=a^3+b^3}}$

$\\\\\\\implies\sf\displaystyle\int \sf \blue{( \sin^3 x + \cos^3 x)}(\sec^2x\;cosec^2x )dx$

$\\\bigstar\;\;\boxed{\sec\theta=\frac{1}{\cos\theta}}\;\;\;\;\;\;\;\;\;\;\;\bigstar\;\;\boxed{cosec\theta=\frac{1}{\sin\theta}}$

$\\\\\\\implies\sf\displaystyle\int \sf ( \sin^3 x + \cos^3 x)(\frac{1}{\cos^2x}\frac{1}{\sin^2x})dx\\\\\implies\sf\displaystyle\int \sf ( \sin^3 x )(\frac{1}{\cos^2x}\frac{1}{\sin^2x})+ (\cos^3 x)(\frac{1}{\cos^2x}\frac{1}{\sin^2x})dx\\\\\implies\sf\displaystyle\int \sf(\frac{\sin x}{\cos^2x})+(\frac{\cos x}{\sin^2x})dx$

We can solve from here in 2 ways.

• 1st method :

$\\\\\\\implies\sf\displaystyle\int \sf(\frac{\sin x}{\cos x}).\frac{1}{\cos x}+(\frac{\cos x}{\sin x}).\frac{1}{\sin x}dx$

$\\\\\bigstar\;\;\boxed{\tan\theta=\frac{\sin\theta}{\cos\theta}}\;\;\;\;\;\;\;\;\;\;\;\bigstar\;\;\boxed{\cot\theta=\frac{\cos\theta}{\sin\theta}}$

$\bigstar\;\;\boxed{\sec\theta=\frac{1}{\cos\theta}}\;\;\;\;\;\;\;\;\;\;\;\bigstar\;\;\boxed{cosec\theta=\frac{1}{\sin\theta}}$

$\\\\\\\implies\sf\displaystyle\int \sf\tan x.\sec x + \cot x.cosec x.dx\\\\\implies\sf\displaystyle\int \sf\tan x.\sec x + \displaystyle\int\cot x.cosec x.dx$

$\bigstar\;\dfrac{d}{dx}\sec x =\tan x.\sec x \\\\\\ \sf\bigstar\;\dfrac{d}{dx}(-cosec x )=\cot x.cosec x$

so,

$\implies \sf \sec x - cosec x + c$

• Method 2:

$\\\\\implies\sf\displaystyle\int \sf(\frac{\sin x}{\cos^2x})+\displaystyle\int(\frac{\cos x}{\sin^2x})dx$

$\\\\\\\implies\sf\displaystyle\int \sf(\frac{-d(\cos x)}{\cos^2x})+\displaystyle\int(\frac{d(\sin x)}{\sin^2x})dx\\\\\implies\sf\displaystyle-\int {\cos x}^{-2}.d(\cos x)+\displaystyle\int{\sin^x}^{-2}.d(\sin x)dx\\\\\implies\sf\displaystyle-\dfrac{{\cos x}^{-2+1}}{-2+1}+\dfrac{{\sin x}^{-2+1}}{-2+1}+c\\\\\implies\sf\displaystyle-\dfrac{{\cos x}^{-1}}{-1}+\dfrac{{\cos x}^{-1}}{-1}+c\\\\\implies\sf\displaystyle \sf\sec x - cosec x+c\\\\\implies\sf\displaystyle \sf\purple{ - cosec x +\sec x+ c}$

Therefore,

$\displaystyle\int \sf (\cos x + \sin x )(1-\frac{1}{2}\sin2x )(\sec^2xcosec^2x )dx =- cosec x + \sec x + c$

• Option c is correct answer.  

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