Question

$\int \frac{x+(arccos3x{)}^2}{\sqrt{1-9x^2}}dx=\frac{1}{k_1}(\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^{k_2})+C$, then $k_1^2+k_2^2=$ (where C is an arbitrary constant)

Answer 1

Let $x=\frac{1}{3}cost$

$\therefore dx=-\frac{1}{3}sintdt$

Substituting $x$ and $dx$ in the given expression,

$\int \frac{x+(co{s}^{-1}3x{)}^2}{\sqrt{1-9x^2}}dx=\int \frac{\frac{1}{3}cost+{\left[co{s}^{-1}(3\times \frac{1}{3}cost)\right]}^2}{\sqrt{1-9\times \frac{1}{9}co{s}^{2}t}}\times -\frac{1}{3}sintdt$

$=\int \frac{\frac{1}{3}cost+t^2}{\sqrt{1-co{s}^{2}t}}\times -\frac{1}{3}sintdt$

$=\int \frac{\frac{1}{3}cost+t^2}{sint}\times -\frac{1}{3}sintdt$

$=\int (-\frac{1}{9}cost-\frac{1}{3}{t}^2)dt$

$=-\frac{1}{9}sint-\frac{1}{9}{t}^3+C$

$=-\frac{1}{9}\{\sqrt{1-9x^2}+(co{s}^{-1}3x{)}^3\}+C$

Hence, $k_1=-9$ and $k_2=3$

$⟹k_1^2+k_2^2=81+9=90$