Mathematics

$$\displaystyle \int \frac{1}{\sqrt{9 -16x^{2}}} dx$$ =


SOLUTION
$$\int { \dfrac { 1 }{ \sqrt { 9-{ 16x }^{ 2 } }  } dx } $$

$$=\int { \dfrac { 1 }{ \sqrt {16\left( \dfrac { 9 }{ 16 } -{ x }^{ 2 } \right)  } dx} } $$

$$=\dfrac{1}{4}\int { \dfrac { 1 }{ \sqrt { \dfrac { 9 }{ 16 } -{ x }^{ 2 } }  } dx } $$

$$=\dfrac{1}{4}\int { \dfrac { 1 }{ \sqrt { { \left( \dfrac { 3 }{ 4 }  \right)  }^{ 2 }-{ x }^{ 2 } }  } dx } $$

$$=\dfrac{1}{4}\sin^{-1}\dfrac{x}{\dfrac34}+c$$              $$\because \int { \dfrac { 1 }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } }  } dx } =\sin ^{ -1 }{ \dfrac { x }{ a } +c } $$

$$=\dfrac{1}{4}\sin^{-1}\dfrac{4x}{3}+c$$

$$=\dfrac{1}{4}\sin^{-1}\dfrac{4x}{3}+c.$$

Hence, the answer is $$\dfrac{1}{4}\sin^{-1}\dfrac{4x}{3}+c.$$
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Subjective Medium Published on 17th 09, 2020
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