Question

Assertion :If $n>1$ then Statement -1: ${\int }_0^{\infty }\frac{dx}{1+x^n}={\int }_0^1\frac{dx}{(1-x^n{)}^{1/n}}$ Reason: Statement -2: ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Put $x^n={\tan }^2\theta \Rightarrow n{x}^{n-1}dx=2\tan \theta {\sec }^2\theta d\theta$
$\Rightarrow dx=\frac{2\tan \theta {\sec }^2\theta }{2{\tan }^{2(n-1)/n}\theta }d\theta$
${\int }_0^{\infty }\frac{dx}{1+x^n}=\frac{2}{n}{\int }_0^{\pi /2}{\tan }^{1-2+2/n}\theta d\theta$
$=\frac{2}{n}{\int }_0^{\pi /2}{\tan }^{2/n-1}\theta d\theta$
Putting $x^n={\sin }^2\theta$ in $R.H.S.$ of statement 1$\Rightarrow n{x}^{n-1}dx=2\sin \theta \cos \theta d\theta \Rightarrow dx=\frac{2\sin \theta \cos \theta }{n{\sin }^{(n-1)/n}\theta }$
$={\int }_0^1\frac{dx}{(1-x^n{)}^{1/n}}=\frac{2}{n}{\int }_0^{\pi /2}\frac{1}{\left({\cos }^{2/n}\theta \right)}{\sin }^{2/n-1}\theta \cos \theta d\theta$
$=\frac{2}{n}{\int }_0^{\pi /2}{\tan }^{2/n-1}\theta d\theta$
Clearly statement 2 is true