Question

${\int }_0^{\pi /4}({\tan }^{n}x+{\tan }^{n-2}x)\frac{d}{dx}\left(3\left\{x\right\}\right)dx$ equals

• A. $\frac{1}{n-1}$
• B. $\frac{2}{n-2}$
• C. $\frac{3}{n-1}$
• D. $\frac{3}{n-2}$

Answer 1

The correct option is C $\frac{3}{n-1}$

${\int }_0^{\frac{\pi }{4}}({\tan }^{n}x+{\tan }^{n-2}x)\left(3\right)dx$
we know $\left\{x\right\}=x-\left[x\right]$
As $0$ to $\frac{\pi }{4},\left[x\right]=0$
$=3{\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}x\left({\sec }^{2}x\right)dx$

$=\frac{3}{n-1}$