Question

If $\int \frac{x+\left(cos^{-1}\,3x\right)^{2}}{\sqrt{1-9x^{2}}}dx=A\sqrt{1-9x^{2}}+B\left(cos^{-1}\,3x\right)^{3}+c,$ 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)

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