Find $\int \frac{dx}{\sqrt{9-4x^{2}}}$

$\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{2x}{3}\right)+C$

CBSE Class 12 · 2020 · Indefinite IntegralsVery Short Answer

Find

\int \frac{d x}{9 - 4 x ^{2}}

Answer: $\frac{1}{2} sin^{- 1} (\frac{2 x}{3}) + C$

Write

9 - 4 x^{2} = 3^{2} - (2 x)^{2}

. With

u = 2 x

d u = 2 d x

and

\int \frac{d u}{3 ^{2} - u ^{2}} = sin^{- 1} \frac{u}{3}

\int \frac{d x}{3 ^{2} - ( 2 x ) ^{2}} = \frac{1}{2} sin^{- 1} (\frac{2 x}{3}) + C .

\frac{1}{2} sin^{- 1} (\frac{2 x}{3}) + C

CBSE Class 12 Board · 2020 · Subjective