Question

The antiderivative of $\frac{x+(co{s}^{-1}3x{)}^2}{\sqrt{1-9x^2}}$ is

• A. $C-\frac{1}{9}[\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^3]$
• B. $C+\frac{1}{9}[\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^2]$
• C. $C-\frac{1}{3}\left[\right(1-9x^2{)}^{3/2}+\left(co{s}^{-1}3x{)}^3\right]$
• D. none of these

Answer 1

The correct option is A $C-\frac{1}{9}[\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^3]$

Let $I=\int \frac{x+(co{s}^{-1}3x{)}^2}{\sqrt{1-9x^2}}dx=\int (\frac{x}{\sqrt{1-9x^2}}+\frac{(co{s}^{-1}3x{)}^2}{\sqrt{1-9x^2}})dx$$=I_1+I_2\dots \left(1\right)$
Where $I_1=\int \frac{x}{\sqrt{1-9x^2}}dx$
Put $1-9x^2=t\Rightarrow -18xdx=dt$
$I_1=-\frac{1}{18}\int \frac{1}{\sqrt{t}}dt=-\frac{1}{18}\frac{\sqrt{t}}{\frac{1}{2}}=-\frac{1}{9}\sqrt{1-9x^2}$
And $I_2=\int \frac{(co{s}^{-1}3x{)}^2}{\sqrt{1-9x^2}}dx$
put $(co{s}^{-1}3x{)}^2=u^2\Rightarrow \frac{-6co{s}^{-1}3x}{\sqrt{1-9x^2}}dx=2udu$
$I_2=-\frac{1}{3}\int u^{2}du=-\frac{1}{9}{u}^3+c=-\frac{1}{9}(co{s}^{-1}3x{)}^3$

From $\left(1\right)$
$I=-\frac{1}{9}[\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^3]+C$