Question

$\int \frac{\sqrt{1-x^2}}{x^4}dx=A\left(x\right)(\sqrt{1-x^2}{)}^m+c$ then $\left(A\right(x){)}^m$ is

• A. $-\frac{1}{27x^9}$
• B. $\frac{1}{27x^9}$
• C. $-\frac{1}{9x^6}$
• D. $-\frac{1}{9x^9}$

Answer 1

The correct option is A $-\frac{1}{27x^9}$

$\int \frac{\sqrt{1-x^2}}{x^4}dx$

substitute $x=\sin u\to dx=\cos udu$

$\int \frac{{\cot }^{2}u}{{\sin }^{2}u}du$

substitute $v=\cot u\to dv=-{\csc }^{2}udu$

$-\int v^{2}dv$

$=-\frac{v^3}{3}$

$=-\frac{{\cot }^{3}u}{3}$

$=-\frac{{\cot }^3\left({\sin }^{-1}x\right)}{3}$

$=-\frac{\sqrt{(1-x^2{)}^3}}{3x^3}+C$

$\therefore m=3$

$A\left(x\right)=-\frac{1}{3x^3}$

$\therefore A(x{)}^m={(-\frac{1}{3x^3})}^3=-\frac{1}{27x^9}$