Math, asked by Anonymous, 1 month ago

\bigstar\sf{Integrate}

{\displaystyle{\int × \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}}

Answers

Answered by mathdude500
8

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

Given integral is

\rm :\longmapsto\:{\displaystyle{\int x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}} \: dx

can be rewritten as

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

To evaluate this integral, we use method of Substitution.

So, Substitute

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

\red{\rm :\longmapsto\: 2x \: dx = dy}

So, on substituting the values, we get

\rm \:  =  \: \dfrac{1}{2}{\displaystyle{\int  \:  \sqrt{\dfrac{a^2-y}{a^2 + y}}}} \: dy

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{1}{2}{\displaystyle{\int  \:  \sqrt{\dfrac{a^2-y}{a^2 + y} \times  \frac{ {a}^{2} - y}{ {a}^{2}  - y} }}} \: dy

\rm \:  =  \: \dfrac{1}{2}\displaystyle\int  \: \frac{ {a}^{2}  - y}{ \sqrt{ {a}^{4}  -  {y}^{2} } } \: dy

\rm \:  =  \: \dfrac{1}{2}\displaystyle\int  \: \frac{ {a}^{2}}{ \sqrt{ {a}^{4}  -  {y}^{2} } } \: dy + \dfrac{1}{2}\displaystyle\int  \: \frac{ - y}{ \sqrt{ {a}^{4}  -  {y}^{2} } } \: dy

\rm \:  =  \: \dfrac{ {a}^{2} }{2}\displaystyle\int  \: \frac{1}{ \sqrt{ {a}^{4}  -  {y}^{2} } } \: dy + \dfrac{1}{4}\displaystyle\int  \: \frac{ -2 y}{ \sqrt{ {a}^{4}  -  {y}^{2} } } \: dy

We know,

\boxed{ \tt{ \: \displaystyle\int  \: \frac{dx}{ \sqrt{ {a}^{2}  -  {x}^{2} } } =  {sin}^{ - 1} \dfrac{x}{a} \: }} \\  \\ and \\  \\ \boxed{ \tt{ \: \displaystyle\int  \: \frac{f'(x)}{ \sqrt{f(x)} }  = 2 \sqrt{f(x)} + c \: }} \\

So, using these, we get

\rm \:  =  \: \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{y}{ {a}^{2} } + \dfrac{1}{4} \times 2 \sqrt{ {a}^{4}  -  {y}^{2} } + c

\rm \:  =  \: \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{y}{ {a}^{2} } + \dfrac{1}{2}  \sqrt{ {a}^{4}  -  {y}^{2} } + c

\rm \:  =  \: \dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{1}{2}  \sqrt{ {a}^{4}  -  {x}^{4} } + c

Hence,

\boxed{ \tt{{\displaystyle{\int x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}} dx=\dfrac{ {a}^{2} }{2} {sin}^{ - 1} \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{1}{2}  \sqrt{ {a}^{4}- {x}^{4} } + 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}

Answered by Anonymous
44

Step-by-step Explanation:

.

 \displaystyle{ \int} \tt  \times  \sqrt{ \dfrac{ {a}^{2}  -  {x}^{2} }{ {a}^{2} +  {x}^{2}  } }

 {\displaystyle{\int }  \tt{ x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}dx}

Can be written as,

  \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ 2x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}dx}

Let's,

  •  \tt  {x}^{2}  = y \\
  • \tt 2x.dx = dy

So,

\tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy\sqrt{\dfrac{a^2-y}{a^2+y}}}}

Simplifying It,

: \implies \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy\sqrt{\dfrac{a^2-y}{a^2+y}}}} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ : \implies \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy\sqrt{\dfrac{a^2-y}{a^2+y} \times  \frac{ {a }^{2} - y }{ {a}^{2}  - y} }}}  \:  \: \\  \\: \implies \tt \dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy\sqrt{\dfrac{(a^2-y) ^{2} }{(a^2) ^{2}  - y ^{2} }}}} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ : \implies \tt \dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{a^2-y }{ \sqrt {(a^2) ^{2}  - y ^{2}} }}}} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ : \implies \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{a^2-y }{ \sqrt {a^{4}  - y ^{2}} }}}}    \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

 : \implies \tt\tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{a^2}{ \sqrt {a^{4}  - y ^{2}} }}}} + \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{-y }{ \sqrt {a^{4}  - y ^{2}} }}}} \\

Can be written as,

 : \implies \tt\tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{a^2}{ \sqrt {a^{4}  - y ^{2}} }}}} + \tt\dfrac{1}{4}  {\displaystyle{\int }  \tt{ dy{\dfrac{-2y }{ \sqrt {a^{4}  - y ^{2}} }}}} \\

As we know that;

 \tt{ \displaystyle \int  \tt\dfrac{dx}{ \sqrt{ {a}^{2}  -  {x}^{2} } } }  =  { \sin }^{ - 1}  \dfrac{x}{a}

 \tt{ \displaystyle \int \tt {\dfrac{ {f}'(x) }{ \sqrt{f(x)} }  = 2 \sqrt{f(x)}  + c} }

Finalising the Integration,

: \implies \tt\tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ dy{\dfrac{a^2}{ \sqrt {a^{4}  - y ^{2}} }}}} + \tt\dfrac{1}{4}  {\displaystyle{\int }  \tt{ dy{\dfrac{-2y }{ \sqrt {a^{4}  - y ^{2}} }}}} \\ \\ : \implies \displaystyle \tt{ \frac{ {a}^{2} }{2} } { \sin}^{ - 1}  \frac{y}{ {a}^{2} }  +  \frac{1}{4}  \times 2 \sqrt{ {a}^{4} -  {y}^{2}  }  + c \:  \:  \:  \:  \:  \:  \:  \\ \\ : \implies \displaystyle \tt{ \frac{ {a}^{2} }{2} } { \sin}^{ - 1}  \frac{y}{ {a}^{2} }  +  \frac{1}{2}  \sqrt{ {a}^{4} -  {y}^{2}  }  + c \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

Substituting real value of y by ,

: \implies \displaystyle \tt{ \frac{ {a}^{2} }{2} } { \sin}^{ - 1}  \frac{ {x}^{2} }{ {a}^{2} }  +  \frac{1}{2}  \sqrt{ {a}^{4} -  {y}^{2}  }  + c

.

.

FINAL ANSWER,

.

  •  \tt\dfrac{1}{2}  {\displaystyle{\int }  \tt{ 2x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}dx} = \displaystyle \tt{ \frac{ {a}^{2} }{2} } { \sin}^{ - 1}  \frac{ {x}^{2} }{ {a}^{2} }  +  \frac{1}{2}  \sqrt{ {a}^{4} -  {y}^{2}  }  + c
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