Mathematics

Find the integral
$$\int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} $$


SOLUTION

Consider the given integral.

  $$ I=\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}} $$

 $$ I=\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}+\dfrac{81}{16}-\dfrac{81}{16}}}} $$

 $$ I=\int{\dfrac{dx}{\sqrt{{{\left( \dfrac{9}{4} \right)}^{2}}-{{\left( 2x-\dfrac{9}{4} \right)}^{2}}}}} $$

 

Put

  $$ t=2x-\dfrac{9}{4} $$

 $$ \dfrac{dt}{2}=dx $$

 

Therefore,

  $$ I=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{{{\left( \dfrac{9}{4} \right)}^{2}}-{{t}^{2}}}}} $$

 $$ I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{t}{\dfrac{9}{4}} \right)+C $$

 

Put the value of t in above expression, we get

  $$ I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4\left( 2x-\dfrac{9}{4} \right)}{9} \right)+C $$

 $$ I=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{\left( 8x-9 \right)}{9} \right)+C $$

 

Hence, the value of this integral is $$\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{\left( 8x-9 \right)}{9} \right)+C$$.

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Subjective Medium Published on 17th 09, 2020
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