Question

Find ${\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{(\sec x+\tan x{)}^n}$, where $(n>2)$

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

Answer 1

Answer: (B)

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

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{{\left(\sec x+\tan x\right)}^n}dx$
Substitute $\sec x+\tan x=t\Rightarrow \left(\sec x\tan x+{\sec }^{2}x\right)dx=dt$
$\Rightarrow \sec xdx=\frac{dt}{t}$
$\therefore \sec x-\tan x=\frac{1}{t}\Rightarrow \sec x=\frac{1}{2}(t+\frac{1}{t})$
$\therefore I=\frac{1}{2}{\int }_1^{\infty }\frac{(t-\frac{1}{t})\frac{dt}{t}}{t^n}=\frac{1}{2}{\int }_1^{\infty }(\frac{1}{t^n}+\frac{1}{t^{+n-12}})dt$
$=\frac{1}{2}{[\frac{1}{(1-n)t^{n-1}}+\frac{1}{(n+1)t^{n+1}}]}_0^{\infty }=\frac{n}{n^2-1}$