Question

$\bf \: evaluate \: the \: definate \: integral \ \: : \\ \displaystyle \rm = \int \frac{1}{ \sqrt{4 - {x}^{2} } } \: dx$

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Answer 1

To Evalute :

${\displaystyle \rm \int \frac{1}{ \sqrt{4 - {x}^{2} } } \: dx}$

Let's Assume :

$\tt{ \bull \frac{x}{2} = sin \:u}$

$\tt{ \bull\frac{dx}{2} = cos \: u \times du}$

$\implies \tt{dx = 2cos(u)du}$

$\tt{ \bull x = 2sin(u)}$

Let's StarT :

We can solve this types of integral by using the trigonometric fluctuations assuming them.

Now, as we assumed earlier (see the let's Assume part ) , we will substitute the values here .

we will also use the identity:

Sin²A+Cos²A = 1

or, Cos²A = 1-sin²A

${\displaystyle \rm \int \frac{1}{ \sqrt{4 - {x}^{2} } } \: dx}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{ \sqrt{4(1 - \frac{ {x}^{2} }{4}) } }dx}$

$\tt{ \displaystyle \rm = \int \frac{1}{2 \sqrt{1 - ( \frac{x}{2} {)}^{2} }}dx}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{2 \sqrt{1 - (sin \: u {)}^{2} } } \: 2cos(u)du}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{ \cancel{2} \sqrt{1 - si {n}^{2} u } } \: \cancel{2}cos(u)du}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{ \sqrt{co {s}^{2}(u) } } \: cos(u)du}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{cos(u)} \: cos(u)du}$

$\tt{ \displaystyle \rm = \int \: \frac{1}{ \cancel{cos(u)}} \: \: \cancel{cos(u)}du}$

$\tt{ \displaystyle \rm = \int \: 1du}$

$\tt{ \rm = u + c}$

Here, We got the result as u+c, so as we know,

if x/2 is sin u , then u is arcsin x/2

$\tt{ = si {n}^{ - 1}( \frac{x}{2} ) + c}$

$\tt{ = arcsin( \frac{x}{2}) + c}$

Hence,

$\blue{\underline{\boxed{ \therefore\displaystyle \rm \int \frac{1}{ \sqrt{4 - {x}^{2} } } \: dx = arcsin (\frac{x}{2}) + c}}}$

Hope it help you!

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{ \sqrt{4 - {x}^{2} } }$

can be rewritten as

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

We know,

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

So, using this result, we get

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

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{4 - {x}^{2} } } \rm \:  =  \: {sin}^{ - 1}\dfrac{x}{2} + c \: }}$

Note :-

Let we derive an expression for integration of

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

To evaluate this integral, we use method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\:x = a \: siny}$

$\purple{\rm :\longmapsto\:dx = a \: cosy \: dy}$

So, on substituting the values, we get

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

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

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

$\rm \:  =  \: \displaystyle\int\rm \frac{a \: cosy \: dy}{ a \: cosy }$

$\rm \:  =  \: \displaystyle\int\rm dy$

$\rm \:  =  \: y \: + \: c$

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

Hence,

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

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MORE TO KNOW

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