If d x 16-9 x^2 =A -1(B x )+C, then A+B= - Vidyadip

MCQ

If $\int \frac{d x}{\sqrt{16-9 x^2}}=A \sin -1(B x )+C$, then $A+B=$

A
$\frac{9}{4}$
B
$\frac{19}{4}$
C
$\frac{3}{4}$
✓
$\frac{13}{12}$

✓

Answer

Correct option: D.

$\frac{13}{12}$

Let $I=\int \frac{d x}{\sqrt{16-9 x^2}}
=\frac{1}{3} \int \frac{d x}{\sqrt{(4 / 3)^2-x^2}}$
$=\frac{1}{3} \sin ^{-1} \frac{x}{(4 / 3)}+C$
$=\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)+C $
On comparing above equation with $A \sin ^{-1}(B x)+C$,
we get $A=\frac{1}{3}, B=\frac{3}{4}$
$\Rightarrow A+B=\frac{1}{3}+\frac{3}{4}$
$=\frac{13}{12}$

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