Question

If $\int \frac{x+{\left(co{s}^{-1}3x\right)}^2}{\sqrt{1-9x^2}}dx=A\sqrt{1-9x^2}+B{\left(co{s}^{-1}3x\right)}^3+c,$ where c is integration constant, then the values of A and B are :

• A. $A=-\frac{1}{9},B=-\frac{1}{9}$
• B. $A=-\frac{1}{9},B=\frac{1}{9}$
• C. $A=\frac{1}{9},B=-\frac{1}{9}$
• D. None of these

Answer 1

Answer: (A)

$I=\int \frac{x+{\left(co{s}^{-1}3x\right)}^2}{\sqrt{1-9x^2}}dx$
Put $3x=cos\theta \Rightarrow 3dx=-sin\theta d\theta$
$I=-\frac{1}{3}\int \frac{\frac{cos\theta }{3}+{\theta }^2}{sin\theta }sin\theta d\theta$
$=-\frac{1}{3}\int \left[\frac{1}{3}cos\theta +{\theta }^2\right]d\theta =-\frac{1}{9}sin\theta -\frac{{\theta }^3}{9}+c$
$=-\frac{1}{9}\sqrt{1-9x^2}-\frac{1}{9}{\left(co{s}^{-1}3x\right)}^3+c$
$\therefore A=B=-\frac{1}{9}$