Question

Which of the following definite integrals reduces to $\frac{\pi }{2}$?

• A. $\underset{0}{\overset{\pi }{\int }}\frac{dx}{1+(\sin x{)}^{\cos x}}$
• B. $\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+(\tan x{)}^5}$
• C. $\underset{0}{\overset{\infty }{\int }}\frac{x^2+1}{x^4-x^2+1}dx$
• D. $\underset{0}{\overset{\pi /2}{\int }}(\ln \left(\sec x\right))(e^{\ln \left(\ln 2\right)}{)}^{-1}dx$

Answer 1

Answer: (A), (D)

The correct options are
A $\underset{0}{\overset{\pi }{\int }}\frac{dx}{1+(\sin x{)}^{\cos x}}$
D $\underset{0}{\overset{\pi /2}{\int }}(\ln \left(\sec x\right))(e^{\ln \left(\ln 2\right)}{)}^{-1}dx$

Let $I_1=\underset{0}{\overset{\pi }{\int }}\frac{dx}{1+(\sin x{)}^{\cos x}}\cdots \left(1\right)$
$\therefore I_1=\underset{0}{\overset{\pi }{\int }}\frac{dx}{1+(\sin x{)}^{-\cos x}}\cdots \left(2\right)$
On adding the above equations, we get
$2I_1=\underset{0}{\overset{\pi }{\int }}1dx$
$\Rightarrow I_1=\frac{\pi }{2}$

Let $I_2=\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+(\tan x{)}^5}\cdots \left(1\right)$
$\therefore I_2=\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+(\cot x{)}^5}\cdots \left(2\right)$
On adding the above equations, we get
$2I_2=\underset{0}{\overset{\pi /2}{\int }}1dx$
$\Rightarrow I_2=\frac{\pi }{4}$

Let $I_3=\underset{0}{\overset{\infty }{\int }}\frac{x^2+1}{x^4-x^2+1}dx$
$=\underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1}dx$
Put $x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt$
So, $I_3=\underset{-\infty }{\overset{\infty }{\int }}\frac{dt}{t^2+1}dx=2{\left({\tan }^{-1}t\right)}_0^{\infty }=\pi$
$\Rightarrow I_3=\pi$

Let $I_4=\underset{0}{\overset{\pi /2}{\int }}\frac{\ln \left(\sec x\right)}{\ln 2}dx$
$=-\frac{1}{\ln 2}\underset{0}{\overset{\pi /2}{\int }}\ln \left(\cos x\right)dx$
$=-\frac{1}{\ln 2}\left(\frac{-\pi }{2}\ln 2\right)=\frac{\pi }{2}$