Question

❥Anushka࿐​​
Q1.integration of √ ( a² - x² ) dx is?​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \sqrt{ {a}^{2} - {x}^{2} } \: dx$

To solve this integral, we use method of Substitution.

$\rm :\longmapsto\:Put \: x = a \: siny$

$\rm :\longmapsto\:dx = a \: cosy \: dy$

and

$\rm :\longmapsto\:y = {sin}^{ - 1}\dfrac{x}{a}$

So, given integral can be rewritten as

$\rm \:  =  \:  \: \displaystyle\int\sf \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} y} \:( a \: cosy )\: dy$

$\rm \:  =  \:  a\: \displaystyle\int\sf \sqrt{ {a}^{2}(1 - {sin}^{2} y)} \: cosy \: dy$

$\rm \:  =  \: a \: \displaystyle\int\sf \sqrt{ {a}^{2}({cos}^{2} y)} \: cosy \: dy$

$\rm \:  =  \: a \: \displaystyle\int\sf a \: cosy \: cosy \: dy$

$\rm \:  =  \:  {a}^{2} \: \displaystyle\int\sf {cos}^{2}y \: dy$

$\rm \:  =  \:  \dfrac{ {a}^{2} }{2} \: \displaystyle\int\sf 2{cos}^{2}y \: dy$

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \: \displaystyle\int\sf (1 + cos2y) \: dy$

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \: \bigg(y + \dfrac{sin2y}{2} \bigg) + c$

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \: \bigg(y + \dfrac{2siny \: cosy}{2} \bigg) + c$

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \: \bigg(y + siny \: cosy \bigg) + c$

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \: \bigg(y + siny \: \sqrt{1 - {sin}^{2} y} \bigg) + c$

On substituting back the values we get

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

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

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

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

$\rm \:  =  \:  \dfrac{a {}^{2} }{2} \:{sin}^{ - 1}\dfrac{x}{a} + \dfrac{x}{ 2 }\: \sqrt{{a}^{2} - {x}^{2} }+ c$

Hence,

$\boxed{ \rm{ \displaystyle\int\rm \sqrt{ {a}^{2} - {x}^{2}} \: dx =  \:  \dfrac{a {}^{2}}{2} \:{sin}^{ - 1}\dfrac{x}{a} + \dfrac{x}{ 2 }\: \sqrt{{a}^{2} - {x}^{2} }+ c}}$

Formula Used:-

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{ \bf{ \: 1 + cos2x = {2cos}^{2}x}}$

$\boxed{ \bf{ \: sin2x = 2sinx \: cosx}}$

$\boxed{ \bf{ \: \displaystyle\int\sf k \: dx = x + c}}$

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

Answer 2

Answer:

$\orange{ \sf\displaystyle \int \sqrt{ {a}^{2} - {x}^{2} } = \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c}$

Step-by-step explanation:

$\bf \: given \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ \sf substitute \\ \sf\: \: x = asin \theta \: \implies \: \theta = {sin}^{ - 1} \frac{x}{a} \: \: \: and \\ \sf \: \pink{dx = acos\theta \: d\theta \: } \\ \bf \: now \\ \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ = \sf \int \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} \theta} \: \: acos\theta \: d\theta \\ = \sf \int \sqrt{ {a}^{2} (1 - {sin}^{2} \theta)} \: \: acos\theta \: d\theta \\ = \sf \int \: a \sqrt{ {1 - {sin}^{2} \theta} } \: a cos\theta \: d\theta \\ = \sf \int {a}^{2} . {cos}^{2} \theta \: d\theta \: \\ = \sf {a}^{2} \int \: \frac{1}{2} (1 + cos2\theta) \: d\theta \: \: \: \: \sf \green{\{ \because \: 2 {cos}^{2} \theta = 1 + cos2\theta \}} \\ \sf= \frac{ {a}^{2} }{2} (\theta + \frac{sin2\theta}{2} ) + c \: \: \: \: \: \: \: \ \sf \red{ \{c = integral \: constant \}} \\ \\ \sf = \frac{ {a}^{2} }{2} (\theta + \frac{2sin\theta \: cos\theta}{2} ) + c \\ \\ \sf = \frac{ {a}^{2} }{2}(\theta + sin \theta \: \sqrt{1 - {sin}^{2}\theta } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} ( {sin}^{ - 1} \frac{x}{a} + \frac{x}{a} \sqrt{1 - \frac{ {x}^{2} }{ {a}^{2} } } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{ {a}^{2} }{2} . \frac{x}{ {a}^{2} } \sqrt{ {a}^{2} - {x}^{2} } + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c$