Math, asked by abhijeetvshkrma, 3 months ago

Solve this integral asap

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Answered by Flaunt
173

\sf\huge\bold{Solution}

Step by step explanation:

 \dfrac{1}{ \sqrt{\bigg(x -  a\bigg)\bigg(x - b\bigg)} }

 =  \dfrac{1}{ \sqrt{x\bigg(x - b\bigg) - a\bigg(x - b\bigg)} }

\sf \longmapsto\displaystyle\int \dfrac{1}{ \sqrt{ {x}^{2} - bx - ax + ab } } dx

\sf \longmapsto\displaystyle\int \dfrac{1}{ \sqrt{ {x}^{2} - x\bigg(a + b\bigg) + ab } } dx

\sf \longmapsto\displaystyle\int \dfrac{1}{ \sqrt{ {\bigg(x -  \dfrac{a + b}{2}\bigg )}^{2} -  {\bigg( \dfrac{a + b}{2} \bigg)}^{2}  + ab }   } dx

\sf \longmapsto\displaystyle\int \dfrac{1}{ \sqrt{ {\bigg(x -  \dfrac{a + b}{2}\bigg) }^{2}  -  \dfrac{ {a}^{2}  -  {b}^{2} - 2ab + 4ab }{4} } } dx

\sf \longmapsto\displaystyle\int  \dfrac{1}{ \sqrt{ {\bigg(x -  \dfrac{a + b}{2} \bigg)}^{2} -  {\bigg( \dfrac{a - b}{2}\bigg) }^{2}  } } dx

\sf \longmapsto\displaystyle\int \dfrac{1}{ \sqrt{ {\bigg(x -  \dfrac{a + b}{2}\bigg) }^{2} -   {\bigg( \dfrac{a - b}{2}\bigg) }^{2}  } } dx

\sf \longmapsto\displaystyle\int \: log\bigg |x +  \sqrt{ {x}^{2}  -  {a}^{2} }\bigg |  + C

\sf \longmapsto\displaystyle\int \bigg|x -  \dfrac{a + b}{2}  +  \sqrt{ {\bigg(x -  \dfrac{a + b}{2} \bigg)}^{2} -  {\bigg( \dfrac{a - b}{2}\bigg )}^{2}  }\bigg |  + C

\sf \longmapsto\displaystyle\int \: log\bigg |x -  \dfrac{a + b}{2}  +  \sqrt{ {x}^{2}  +  {\bigg( \dfrac{a + b}{2}\bigg) }^{2}  - 2\bigg(x\bigg)\bigg( \dfrac{a + b}{2} \bigg) -  {\bigg( \dfrac{a - b}{2} \bigg)}^{2} } \bigg|  + C

\sf \longmapsto\displaystyle\int \: log\bigg |x -  \dfrac{a + b}{2}  +  \sqrt{ {x}^{2} - 2\bigg(x\bigg)\bigg( \dfrac{a + b}{2}\bigg ) +  {\bigg( \dfrac{a + b}{2}\bigg) }^{2}   -  {\bigg( \dfrac{a - b}{2}\bigg )}^{2} } \bigg|  + C

\sf \longmapsto\displaystyle\int \: log\bigg |x -  \dfrac{a + b}{2}  \sqrt{ {x}^{2} - x\bigg(a + b\bigg) +  \dfrac{ {a}^{2}  +  {b}^{2}  + 2ab}{4}  -  \dfrac{ {a}^{2}  +  {b}^{2} - 2ab }{4}  } \bigg|  + C

\sf \longmapsto\displaystyle\int \: log\bigg |x -  \dfrac{a + b}{2}  +  \sqrt{ {x}^{2}  - x\bigg(a + b\bigg) +  \dfrac{2ab}{4}  +  \dfrac{2ab}{4} }\bigg |  + C

\sf \longmapsto\displaystyle\int \: log \bigg|x -  \dfrac{a + b}{2} +  \sqrt{ {x}^{2} -x\bigg( a  +  b \bigg)+  \dfrac{4ab}{4}  } \bigg |  + C

\sf \longmapsto\displaystyle\int \: log \bigg|x -  \dfrac{a + b}{2}  +  \sqrt{ {x}^{2}  - ax - bx + ab} \bigg|  + C

\sf \longmapsto\displaystyle\int \: log\bigg |x -  \dfrac{a + b}{2} +  \sqrt{x\bigg(x - a\bigg) - b\bigg(x - a\bigg)} \bigg |  + C

  \sf= log\bigg |x -  \dfrac{a + b}{2}  +  \sqrt{(x - a)(x - b)} \bigg|  + C


IdyllicAurora: Nice :)
abhijeetvshkrma: Thanks bhai ❤️
Flaunt: thanks :)
Asterinn: Great !
prince5132: Superb !
Flaunt: Thanks all !
Anonymous: Nicee
Answered by amansharma264
9

EXPLANATION.

\sf \:  \implies \:  \int \dfrac{1}{ \sqrt{(x - a)(x - b)} } dx.

\sf \:  \implies \:  \int \dfrac{1}{  \sqrt{( {x}^{2} - xb - xa + ab) }  }dx.

\sf \:  \implies \:  \int \dfrac{1}{ \sqrt{( {x}^{2}  - (a + b)x  + ab} } dx.

Multiply and divide by 2 in equation, we get.

\sf \:  \implies \:   \int \: \dfrac{1}{ \sqrt{( {x}^{2}  - 2(x) \bigg ( \dfrac{a + b}{2}  \bigg) + ab} }dx.

Change equation into perfect square, we get.

\sf \:  \implies \:  \int  \dfrac{1}{ \sqrt{( {x}^{2}  - 2x \bigg( \dfrac{a + b}{2} \bigg) +  \bigg( \dfrac{a + b}{2}  \bigg) {}^{2}   -  \bigg( \dfrac{a + b}{2}  \bigg) {}^{2}  + ab} } dx. \\  \\  \\  \sf \:  \implies \:  \int \:  \dfrac{1}{ \sqrt{ \bigg(x -  \dfrac{a + b}{2}  \bigg) {}^{2}  -  \bigg( \dfrac{a + b}{2}  \bigg) {}^{2}  + ab} } dx. \\  \\  \\ \sf \:  \implies \: factorise \: the \: equation \: we \: get \\  \\  \\ \sf \:  \implies \:  \int \dfrac{1}{ \sqrt{ \bigg(x -  \dfrac{a + b}{2}  \bigg) {}^{2}  -  \bigg( \dfrac{ {a}^{2}  +  {b + 2ab}^{2} }{2}   + ab \bigg)} } dx.

\sf \:  \implies \:  \int \:  \dfrac{1}{ \sqrt{ \bigg(x -  \dfrac{a + b}{2}  \bigg) {}^{2}   +   \bigg( \dfrac{ { - a}^{2}  -  {b}^{2}  - 2ab + 4ab}{4}  \bigg)} } dx. \\  \\  \\ \sf \:  \implies \:  \int \:  \dfrac{1}{ \sqrt{ \bigg(x -  \dfrac{a + b}{2} \bigg) {}^{2}   -  \bigg( \dfrac{ {a}^{2}  +  {b}^{2}  - 2ab}{4}  \bigg)} } dx. \\  \\  \\ \sf \:  \implies \:  \int \:  \dfrac{1}{ \sqrt{ \bigg(x -  \dfrac{a + b}{2} \bigg) {}^{2}   -  \bigg( \dfrac{a - b}{2} \bigg) {}^{2}  } } dx.

\sf \:  \implies \: as \: we \: know \: that \\  \\  \\ \sf \:  \implies \:  \int \:  \frac{dx}{ \sqrt{ {x}^{2}  -  {a}^{2} } }  =  log \bigg |x +  \sqrt{ {x}^{2}  -  {a}^{2} }   \bigg|  + c

\sf \implies \int log \bigg|x - \dfrac{a + b}{2}+\sqrt{x^{2} - 2(x)\bigg(\dfrac{a + b}{2}\bigg)+ \bigg(\dfrac{a + b}{2} \bigg)^{2} - \bigg(\dfrac{a - b}{2} \bigg)^{2}  }  \bigg| + c

\sf \implies \int log \bigg| x - \dfrac{a + b}{2} + \sqrt{x^{2} - x(a + b) + \dfrac{a^{2}+b^{2}  +2ab}{4}- \dfrac{a^{2} + b^{2}-2ab }{4}  } \bigg| + c

\sf \implies log \bigg| x - \dfrac{a + b}{2} + \sqrt{x^{2} - x(a + b) + \dfrac{2ab}{4} + \dfrac{2ab}{4}  } \bigg| + c

\sf \implies log \bigg | x - \dfrac{a + b}{2} +\sqrt{x^{2} - x(a + b) +\dfrac{4ab}{4}  } \bigg| + c

\sf \implies log \bigg| x - \dfrac{a + b}{2} +\sqrt{x^{2} - x(a + b) + ab} \bigg| + c

\sf \implies log \bigg| x - \dfrac{a + b}{2} +\sqrt{(x - a)(x - b)} \bigg| + c


prince5132: Great !
Anonymous: Awesome as always (:
IdyllicAurora: Cool !!
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