Question

If $\left[t\right]$ denotes the greatest integer $\le t,$ then the value of $\underset{0}{\overset{\pi /8}{\int }}[\sec x+2[\tan x+3[\cos x+4[\sin x\left]\right]]]dx$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{10}$

Answer 1

The correct option is C $\frac{\pi }{8}$

$\underset{0}{\overset{\pi /8}{\int }}[\sec x+2[\tan x+3[\cos x+4[\sin x\left]\right]]]dx$
Let $[\tan x+3[\cos x+4[\sin x\left]\right]]=k$
Then, $k\in Z$
We know $[x+n]=\left[x\right]+n,n\in Z$

Now, $[\sec x+2[\tan x+3[\cos x+4[\sin x\left]\right]]]$
$=\left[\sec x\right]+2[\tan x+3[\cos x+4[\sin x\left]\right]]=\left[\sec x\right]+2\left(\right[\tan x]+3[\cos x+4[\sin x\left]\right])=\left[\sec x\right]+2\left[\tan x\right]+6[\cos x+4[\sin x\left]\right]=\left[\sec x\right]+2\left[\tan x\right]+6\left[\cos x\right]+24\left[\sin x\right]$
So, $\underset{0}{\overset{\pi /8}{\int }}[\sec x+2[\tan x+3[\cos x+4[\sin x\left]\right]]]dx$
$=\underset{0}{\overset{\pi /8}{\int }}\left[\sec x\right]dx+\underset{0}{\overset{\pi /8}{\int }}2\left[\tan x\right]dx+\underset{0}{\overset{\pi /8}{\int }}6\left[\cos x\right]dx+\underset{0}{\overset{\pi /8}{\int }}24\left[\sin x\right]dx=\underset{0}{\overset{\pi /8}{\int }}1\cdot dx+0+0+0=\frac{\pi }{8}$