Question

Given that $\int \frac{\sin ^3 x-x \sin ^2 x \cdot \cos x}{\cos ^3 x} d x=f(x)+C$, where $C$ is constant of integration. If $f (\pi)=\frac{\pi^2}{2}$, then the value of $[ f (2 \pi)]$ equals
[Note: [k] denotes greatest integer function less than or equal to k.]

• A. 19
• B. 39
• C. 40
• D. 18

Answer 1

Answer: (A)

$\int\left(\tan ^3 x-x \tan ^2 x\right) d x=\int \tan ^2 x(\tan x-x) d x$
Let $\tan x-x=t \Rightarrow\left(\sec ^2 x-1\right) d x=d t \Rightarrow \tan ^2 x d x=d t$
$\int t d t =\frac{ t ^2}{2}+ C =\frac{(\tan x - x )^2}{2}+ C$
$f ( x )=\frac{(\tan x - x )^2}{2} $
$( f (2 \pi))=2 \pi^2 \approx 19.73 $
$\therefore[ f (2 \pi)]=19 .$