Question

Question
$\rm \implies \: for \: {x}^{2} \not = \: n\pi + 1,n \in N \: , \: the \: intergral$
$\implies \rm \int x \sqrt{ \frac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)} } dx \:$

is equal to

$\rm \: a) \dfrac{1}{2} log_{e} | \sec( {x}^{2} - 1 ) | + c$
$\rm b) log_{e} \bigg| \sec \bigg( \dfrac{ {x}^{2} - 1 }{2} \bigg ) \bigg | + c$
$\rm \: c) log_{e} \bigg| \dfrac{1}{2} \sec( {x}^{2} - 1) \bigg| + c$
$\rm d) \dfrac{1}{2} log_{e} \bigg | \sec ^{2} \bigg( \dfrac{ {x}^{2} - 1 }{2} \bigg ) \bigg | + c$

(2019 Main, 9Jan I )

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Answer 1

Question :

$\bullet\ \; \displaystyle \sf \red{ \int x\ \sqrt{\dfrac{2\ sin(x^2-1)-sin2(x^2-1)}{2\ sin(x^2-1)+sin2(x^2-1)}}\ dx }$

For x² ≠ nπ + 1 , n ∈ N

$\sf a)\ \dfrac{1}{2} \log _e |sec(x^2-1)|+c$

$\sf b)\ \log _e \Bigg| sec^2\bigg( \dfrac{x^2-1}{2}\bigg) \Bigg| +c$

$\sf c)\ \log _e \Bigg| \dfrac{1}{2}\ sec(x^2-1) \Bigg| +c$

$\sf d)\ \dfrac{1}{2} \log _e \Bigg| sec^2 \bigg( \dfrac{x^2-1}{2}\bigg) \Bigg|+c$

Solution :

$\\ \displaystyle \sf \int x\ \sqrt{\dfrac{2\ sin(x^2-1)-sin2(x^2-1)}{2\ sin(x^2-1)+sin2(x^2-1)}}\ dx$

$\bullet\ \; \sf \pink{sin2 \theta =2 sin\ \theta cos\ \theta}$

$\\ \to \displaystyle \sf \int x\ \sqrt{\dfrac{2\ sin(x^2-1)-2\ sin(x^2-1)\ cos(x^2-1)}{2\ sin(x^2-1)+2\ sin(x^2-1)\ cos(x^2-1)}}\ dx$

$\\ \to \displaystyle \sf \int x\ \sqrt{\dfrac{2\ sin(x^2-1) \big[1-cos(x^2-1) \big]}{2\ sin(x^2-1) \big[ 1+cos(x^2-1) \big]}}\ dx$

$\\ \to \displaystyle \sf \int x\ \sqrt{\dfrac{1-cos(x^2-1)}{1+cos(x^2-1)}}\ dx$

$\bullet\ \; \sf \pink{1-cos2 \theta =2sin^2 \theta}$

$\implies \sf \pink{1-cos\ \theta =2sin^2 \dfrac{\theta}{2}}$

$\bullet\ \; \sf \pink{1+cos2\ \theta =2cos^2 \theta}$

$\implies \sf \pink{1+cos\ \theta =2cos^2 \dfrac{\theta}{2}}$

$\\ \to \displaystyle \sf \int x\ \sqrt{\dfrac{2\ sin^2\big( \frac{x^2-1}{2}\big)}{2\ cos^2 \big( \frac{x^2-1}{2}\big)}}\ dx$

$\\ \to \displaystyle \sf \int x\ \sqrt{\dfrac{ sin^2\big( \frac{x^2-1}{2}\big)}{ cos^2 \big( \frac{x^2-1}{2}\big)}}\ dx$

$\\ \to \displaystyle \sf \int x\ \sqrt{tan^2\bigg( \frac{x^2-1}{2}\bigg)}\ dx$

$\\ \to \displaystyle \sf \int x\ tan \bigg( \frac{x^2-1}{2}\bigg)\ dx$

★ ════════════════════ ★

Let's use substitution ,

$\mapsto\ \sf \pink{u=\dfrac{x^2-1}{2}}$

$\mapsto \sf \pink{du=\dfrac{2x}{2}\ dx}$

$\mapsto \sf \pink{du=x\ dx}$

★ ════════════════════ ★

$\\ \to \displaystyle \sf \int tan\ u\ du$

$\to \sf \log_e (sec\ u)+c$

$\\ \displaystyle \sf \green{ \leadsto \log_e \Bigg| sec \bigg( \dfrac{x^2-1}{2}\bigg) \Bigg|+c}$

Option (b)

Answer 2

$\large\underline\blue{\bold{Formula \: used:- }}$

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$\bf \:(1). \: sin2x = 2sinx \: cosx$

$\bf \:(2). \: 1 - cosx = {2sin}^{2} \dfrac{x}{2}$

$\bf \:(3). \: 1 + cosx = {2cos}^{2} \dfrac{x}{2}$

$\bf \:(4). \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\bf \:(5). \: \int \: tanx = log_e|secx| + c$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \:Let \: S = \rm \int x \sqrt{ \dfrac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)} } dx \: \sf \:  ⟼(1)$

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☆ Let us consider

$\bf \:\dfrac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)}\sf \:  ⟼ \: (2)$

$\sf \:  Let \: {x}^{2} - 1 = t$

☆ So, equation (2) can be rewritten as

$\sf \:  \dfrac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)} = \dfrac{2sint - sin2t}{2sint + sin2t}$

$\sf \:  = \dfrac{2sint - 2sintcost}{2sint + 2sintcost}$

$\sf \:  = \dfrac{2sint(1 - cost)}{2sint(1 + cost)}$

$\sf \:  = \dfrac{ \cancel{2sint} \: (1 - cost)}{ \cancel{2sint} \: (1 + cost)}$

$\sf \:  = \dfrac{ {2sin}^{2}\dfrac{t}{2} }{2 {cos}^{2} \dfrac{t}{2} }$

$\sf \:  = \dfrac{ { \cancel2 \: sin}^{2}\dfrac{t}{2} }{ \cancel2 \: {cos}^{2} \dfrac{t}{2} }$

$\sf \:  = {tan}^{2} \dfrac{t}{2} \sf \:$

$\sf \:  = {tan}^{2} \bigg(\dfrac{ {x}^{2} - 1 }{2} \bigg)$

$\bf\implies \:\dfrac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)} = {tan}^{2} \bigg(\dfrac{ {x}^{2} - 1 }{2} \bigg)$

☆ So, now equation (1) can be rewritten as

$\bf \:S = \rm \int x \sqrt{ \dfrac{2 \sin( {x}^{2} - 1) - \sin2( {x}^{2} - 1) }{2 \sin( {x}^{2} - 1) + \sin2( {x}^{2} - 1)} } dx$

$\bf \:S = \int \: x \sqrt{ {tan}^{2} \bigg(\dfrac{ {x}^{2} - 1 }{2} \bigg)} dx$

$\sf \: S = \int \: x \: {tan} \bigg(\dfrac{ {x}^{2} - 1 }{2} \bigg)dx\sf \:  ⟼(4)$

$\sf \:  Put \: {x}^{2} - 1 = y\sf \:  ⟼ \: (5)$

☆ Differentiate w. r. t. x both sides, we get

$\sf \:  \dfrac{d}{dx} ( {x}^{2} - 1) = \dfrac{dy}{dx}$

$\sf \:  ⟼2x = \dfrac{dy}{dx}$

$\sf \:  ⟼x \: dx = \dfrac{1}{2} dy\sf \:  ⟼(6)$

☆ On substituting equation (5) and (6) in equation (4), we get

$\sf \:  S = \int \: (tan \dfrac{y}{2}) \: \dfrac{dy}{2}$

$\sf \:  S = \dfrac{1}{2} \int \: (tan \dfrac{y}{2}) \: dy$

$\bf\implies \:S = \dfrac{1}{2} \dfrac{log_e |sec \frac{y}{2} | }{\dfrac{1}{2} } + c$

$\bf\implies \:S = log_e |sec\dfrac{1}{2}( {x}^{2} - 1) | + c$

$\bf\implies \:S = log_{e} \bigg| \sec \bigg( \dfrac{ {x}^{2} - 1 }{2} \bigg ) \bigg | + c$

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$\large{\boxed{\boxed{\bf{Option \: (b) \: is \: correct}}}}$