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
Questions 203525
Subjects 9
Chapters 126
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