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❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ

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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {sin}^{ - 1} \sqrt{x} - {cos}^{ - 1} \sqrt{x} }{ {sin}^{ - 1} \sqrt{x} + {cos}^{ - 1} \sqrt{x} } \: dx$

We know,

$\red{\boxed{\tt{ {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} \: }}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{ - 1} \sqrt{x} - \bigg[\dfrac{\pi}{2} - {sin}^{ - 1} \sqrt{x} \bigg]}{\dfrac{\pi}{2}} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{2 {sin}^{ - 1} \sqrt{x} - \dfrac{\pi}{2}}{\dfrac{\pi}{2}} \: dx$

$\rm \:  =  \: \dfrac{4}{\pi}\displaystyle\int\rm {sin}^{ - 1} \sqrt{x} dx - \displaystyle\int\rm 1 \: dx$

$\rm \:  =  \: \dfrac{4}{\pi}I \: - \: x \: + \: c - - - (1)$

where,

$\red{\rm :\longmapsto\:\displaystyle\int\rm {sin}^{ - 1} \sqrt{x} \: dx \: }$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\: \sqrt{x} = siny}$

$\red{\rm :\longmapsto\:x = {sin}^{2}y}$

$\red{\rm :\longmapsto\:dx = 2siny \: cosy \: dy}$

$\red{\rm :\longmapsto\:dx \: = \: sin2y \: dy \: }$

So, above integral can be rewritten as

$\rm :\longmapsto\:I = \displaystyle\int\rm {sin}^{ - 1}(siny) \: sin2y \: dy$

$\rm :\longmapsto\:I = \displaystyle\int\rm y \: sin2y \: dy$

Now, using integration by parts, we have

$\rm :\longmapsto\:I = y\displaystyle\int\rm sin2ydy - \displaystyle\int\rm \bigg[\dfrac{d}{dy} y\displaystyle\int\rm sin2y \: dy\bigg]dy$

$\rm :\longmapsto\:I = - \dfrac{ycos2y}{2} + \displaystyle\int\rm \dfrac{cos2y}{2} \: dy$

$\rm :\longmapsto\:I = - \dfrac{ycos2y}{2} + \dfrac{sin2y}{4}$

$\rm :\longmapsto\:I = - \dfrac{y(1 - {2sin}^{2} y)}{2} + \dfrac{2 \: siny \: cosy}{4}$

$\rm :\longmapsto\:I = - \dfrac{ {sin}^{ - 1} \sqrt{x} (1 - 2x)}{2} + \dfrac{ \: \sqrt{x} \: \sqrt{1 - {sin}^{2} y} }{2}$

$\rm :\longmapsto\:I = - \dfrac{ {sin}^{ - 1} \sqrt{x} (1 - 2x)}{2} + \dfrac{ \: \sqrt{x} \: \sqrt{1 - x} }{2}$

On substituting the value of I in equation (1), we get

$\rm =\dfrac{4}{\pi}\bigg[- \dfrac{ {sin}^{ - 1} \sqrt{x} (1 - 2x)}{2} + \dfrac{ \: \sqrt{x} \: \sqrt{1 - x} }{2}\bigg] - x + c$

$\rm =\dfrac{2}{\pi}\bigg[- {sin}^{ - 1} \sqrt{x} (1 - 2x) + \sqrt{x} \: \sqrt{1 - x}\bigg] - x + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {sin}^{ - 1} \sqrt{x} - {cos}^{ - 1} \sqrt{x} }{ {sin}^{ - 1} \sqrt{x} + {cos}^{ - 1} \sqrt{x} } \: dx$

$\rm =\dfrac{2}{\pi}\bigg[- {sin}^{ - 1} \sqrt{x} (1 - 2x) + \sqrt{x} \: \sqrt{1 - x}\bigg] - x + c$

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Formula Used :-

Integration by parts

$\rm :\longmapsto\:\displaystyle\int\rm \: uvdx = u\displaystyle\int\rm v \: dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx} u\displaystyle\int\rm v \: dx\bigg]dx$

where, u and v are chosen according to the word ILATE.

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

$\boxed{\tt{ \frac{d}{dx} \: x \: = \: 1 \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$