First slide

Introduction to integration

Question

If $\int \frac{2 x - \sqrt{\sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x = C - 2 \sqrt{1 - x^{2}} -$ $\frac{2}{3} \sqrt{f (x)}$ then $f (x)$ is equal to

Easy

A

$\sin^{- 1} x$
B

$2 \sin^{- 1} x$
C

${(\sin^{- 1} x)}^{3}$
D

$3 {(\sin^{- 1} x)}^{3}$

Solution

$The integrand is equal to \frac{2 x}{\sqrt{1 - x^{2}}} - \frac{\sqrt{\sin^{- 1} x}}{\sqrt{1 - x^{2}}} .$
$The first part is of form \frac{- g^{'} (x) ∣}{\sqrt{g (x)}} where g (x) = (1 - x^{2})$
and other part is of the form $\sqrt{h (x)} h^{'} (x), h (x) = \sin^{- 1} x$
$Hence the required integral$
$C - 2 \sqrt{1 - x^{2}} - \frac{2}{3} {(\sin^{- 1} x)}^{3 / 2} . Thus f (x) = {(\sin^{- 1} x)}^{3}$

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