Find $\int \frac{\sin^{-1}x}{\left(1-x^{2}\right)^{3/2}}\,dx$

$\dfrac{x\,\sin^{-1}x}{\sqrt{1-x^{2}}}+\dfrac{1}{2}\log\left(1-x^{2}\right)+C$

CBSE Class 12 · 2023 · Indefinite IntegralsLong Answer

Find

\int \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} d x

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

Put

t = sin^{- 1} x

, so

x = sin t

and

d t = \frac{d x}{1 - x ^{2}}

. Since

(1 - x^{2})^{3/2} = (1 - x^{2}) 1 - x^{2}

and

1 - x^{2} = cos^{2} t

, the integral becomes

\int \frac{t}{1 - sin ^{2} t} d t = \int t sec^{2} t d t .

Integrate by parts with

u = t

d v = sec^{2} t d t

\int t sec^{2} t d t = t tan t - \int tan t d t = t tan t + lo g ∣ cos t ∣ + C .

Now

tan t = \frac{x}{1 - x ^{2}}

and

lo g ∣ cos t ∣ = \frac{1}{2} lo g (1 - x^{2})

, so

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

CBSE Class 12 Board · 2023 · Subjective