Question

intregal{sininverse(√x/a+x)dx}

Answer 1

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

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

Let assume that

${\rm : \: \longmapsto\:I \: = \: \displaystyle\int\rm {sin}^{ - 1} \sqrt{\dfrac{x}{a + x} } \: dx} - - - - (1)$

To evaluate this integral, we use method of Substitution

${\rm : \: \longmapsto\:Put \: \sqrt{\dfrac{x}{a + x} } \: = \: siny}$

$\rm :\longmapsto\:\dfrac{x}{a + x} = {sin}^{2}y$

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

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

$\rm :\longmapsto\:x(1 - {sin}^{2}y )= a {sin}^{2}y$

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

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

$\red{\rm :\longmapsto\:dx = 2a \: tany \: {sec}^{2}y \: dy}$

So, on substituting all these values in given integral, we get

$\rm :\longmapsto\:I = \displaystyle\int\rm y \: \times \: 2a \: tany \: {sec}^{2}y \: dy$

$\rm \: = \: 2a\displaystyle\int\rm y \times tany \: {sec}^{2}y \: dy$

We know, Integration by Parts,

$\boxed{ \bf{ \: \displaystyle\int\rm u.v \: dx \: = \: u\displaystyle\int\rm vdx - \displaystyle\int\rm \bigg(\dfrac{d}{dx}u \: \displaystyle\int\rm vdx\bigg) dx}}$

So, Here,

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

and

$\red{\rm :\longmapsto\:v = tany \: {sec}^{2}y}$

So, on substituting the values, we get

$\rm \: =2a\bigg[y\displaystyle\int\rm tany {sec}^{2} ydy - \displaystyle\int\rm \bigg(\dfrac{d}{dy}y \: \displaystyle\int\rm tany {sec}^{2}ydy \bigg) dy\bigg]$

Now,

$\red{\rm :\longmapsto\:\displaystyle\int\rm tany \: {sec}^{2}y \: dy = \dfrac{1}{2} {tan}^{2}y}$

and

$\red{\rm :\longmapsto\:\dfrac{d}{dy}y = 1}$

So, using these values, we get

$\rm \: = 2a\bigg[y\dfrac{ {tan}^{2} y}{2} - \displaystyle\int\rm 1 \times \dfrac{ {tan}^{2} y}{2} \: dy\bigg]$

$\rm \: = 2a\bigg[y\dfrac{ {tan}^{2} y}{2} -\dfrac{1}{2} \displaystyle\int\rm ( {sec}^{2} y - 1) \: dy\bigg]$

$\rm \: = ay {tan}^{2}y -a tany + ay + c$

As

$\red{\rm :\longmapsto\:x = a {tan}^{2}y\bf\implies \: {tan}^{2}y = \dfrac{ x}{a}}$

So, on substituting all these values, we get

$\rm \: = \: x {tan}^{ - 1} \sqrt{\dfrac{x}{a} } - \sqrt{ax} + a{tan}^{ - 1} \sqrt{\dfrac{x}{a} } + c$

Hence,

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

is

$\bf \: = \: x {tan}^{ - 1} \sqrt{\dfrac{x}{a} } - \sqrt{ax} + a{tan}^{ - 1} \sqrt{\dfrac{x}{a} } + c$

Note :-

Proof of

$\red{\rm :\longmapsto\:\displaystyle\int\rm tany \: {sec}^{2}y \: dy = \dfrac{1}{2} {tan}^{2}y}$

By method of Substitution,

$\rm :\longmapsto\:tany = t$

$\rm :\longmapsto\: {sec}^{2} y \: dy=d t$

So, given integral reduced to

$\rm \: = \: \displaystyle\int\rm t \: dt$

$\rm \: = \: \dfrac{ {t}^{2} }{2} + c$

$\rm \: = \: \dfrac{ {tan}^{2} t}{2} + c$

Answer 2

Step-by-step explanation:

$\frac{ \tan2t}{2} + c$