Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

The value of $\int\limits_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d\left(x-\frac{[x]}{1 !}+\frac{[x]^2}{2 !}-\frac{[x]^3}{3 !}+... \right)$ where [x] is greatest integer function, is 

$[x \in[0, \pi / 4]]$

Options:

$\frac{1}{n}$

$\frac{1}{n+2}$

$\frac{1}{n-1}$

$\frac{1}{n-2}$

Correct Answer:

$\frac{1}{n-1}$

Explanation:

For $x \in[0, \pi / 4],[x]=0$

∴  $\int\limits_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d\left(x-\frac{[x]}{1 !}+\frac{[x]^2}{2 !}-\frac{[x]^3}{3 !}+...\right)$

$=\int\limits_0^{\pi / 4} \tan ^{n-2} x \sec ^2 x d x=\left[\frac{\tan ^{n-1} x}{n-1}\right]_0^{\pi / 4}=\frac{1}{n-1}$