Question

If $\int \frac{dx}{\sqrt[3]{3{\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x}}=f\left(x\right){\tan }^{-1}\left(\frac{f\left(x\right)}{\sqrt{2}}\right)+c$, then $f\left(x\right)$ is-

• A. Bounded & periodic
• B. Bounded & aperiodic
• C. Unbounded & periodic
• D. Unbounded & aperiodic

Answer 1

Answer: (C)

Unbounded & periodic

$\int \frac{dx}{\sqrt[3]{3{\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x}}$

$=\int \frac{dx}{\sqrt[3]{3({\cos }^{4}x+{\cos }^{2}x{)}^3}}$

$=\int \frac{dx}{{\cos }^{4}x+{\cos }^{2}x}$

$=\int \frac{dx}{{\cos }^{2}x({\cos }^{2}x+1)}$

$=\int \frac{{\sec }^{2}x}{2{\cos }^{2}x-{\cos }^{2}x+1}dx$

$=\int \frac{{\sec }^{2}x}{2{\cos }^{2}x+{\sin }^{2}x}dx$

$=\int \frac{1+{\tan }^{2}x}{{\cos }^{2}x(2+{\tan }^{2}x)}dx$

Put, $\tan x=t⟹{\sec }^{2}xdx=dt⟹\frac{1}{{\cos }^{2}x}dx=dt$

$=\int \frac{1+t^2}{2+t^2}dt$

$=\int \frac{2+t^2-1}{2+t^2}dt$

$=\int (1-\frac{1}{(\sqrt{2}{)}^2+t^2})dt$

$=t-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+c$

$=\tan x-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)+c$

$\therefore f\left(x\right)=\tan x$

Hence, It is Unbounded & periodic.

Option C is correct.