Mathematics

# Solve:$\int \frac{dx}{\sqrt{4-x^{2}}}$

##### SOLUTION
$\int \dfrac{dx}{\sqrt4-x^2} \ \ Let \ x=2\sin \theta$

$\dfrac{dx}{d \theta}=2 \cos \theta$

$=\int \dfrac{dx}{\sqrt{4-4 \sin^2} \theta}$

$=\int \dfrac{2\cos \theta d\theta}{\sqrt{4-4 \sin^2} \theta}$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ \sqrt { 4\left( 1-{ \sin }^{ 2 }\theta \right) } } }$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ \sqrt { 4\cos ^{ 2 }\theta } } }$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ 2\cos { \theta } } }$

$=\int d\theta=\theta$

$\therefore \theta=\dfrac{1}{2}\sin^{-1}x+c.$ Ans

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$