Question

If $f\left(x\right)=\int \frac{dx}{\tan x+\sec x+\cot x+\text{cosec}x}$ and $f\left(0\right)=\frac{5}{2},$ then $\left[f\right(\pi \left)\right]$ is equal to
(where [.] represents greatest integer function)

Answer 1

$f\left(x\right)=\int \frac{dx}{\tan x+\sec x+\cot x+\text{cosec}x}\Rightarrow f\left(x\right)=\int \frac{\sin x\cos xdx}{1+\sin x+\cos x}\Rightarrow f\left(x\right)=\int \frac{\sin xdx}{\sec x+\tan x+1}$
Multiplying and dividing by $1+\tan x-\sec x$
$\Rightarrow f\left(x\right)=\int \frac{\sin x(1+\tan x-\sec x)dx}{(1+\tan x{)}^2-{\sec }^{2}x}\Rightarrow f\left(x\right)=\int \frac{\sin x(1+\tan x-\sec x)dx}{2\tan x}\Rightarrow f\left(x\right)=\frac{1}{2}\int \cos x+\sin x-1dx\Rightarrow f\left(x\right)=\frac{1}{2}[\sin x-\cos x-x]+c\Rightarrow f\left(0\right)=c-\frac{1}{2}=\frac{5}{2}\Rightarrow c=3$
Therefore,
$f\left(\pi \right)=\frac{1}{2}[0+1-\pi ]+3\Rightarrow f\left(\pi \right)=\frac{7-\pi }{2}$
Hence, $\left[f\right(\pi \left)\right]=1$