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Question Number 201517 by sonukgindia last updated on 08/Dec/23

	Answered by Calculusboy last updated on 08/Dec/23

	$S o l u t i o n : l e t θ = \frac{π}{2} - x d θ = - d x$ $w h e n x = \frac{π}{2} θ = 0 a n d w h e n x = 0 θ = \frac{π}{2}$ $I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + {(\frac{1}{t a n x})}^{n}} d x \Leftrightarrow I = \int_{0}^{\frac{π}{2}} \frac{1}{\frac{{(t a n x)}^{n} + 1}{{(t a n x)}^{n}}} d x$ $I = \int_{0}^{\frac{π}{2}} \frac{{(t a n x)}^{n}}{1 + {(t a n x)}^{n}} d x I = \int_{\frac{π}{2}}^{0} \frac{{[t a n (\frac{π}{2} - θ)]}^{n}}{1 + {[t a n (\frac{π}{2} - θ)]}^{n}} (- d θ)$ $N b : t a n (\frac{π}{2} - θ) = c o t θ$ $I = \int_{0}^{\frac{π}{2}} \frac{{(c o t θ)}^{n}}{1 + {(c o t θ)}^{n}} d θ (c h a n g i n g o f v a r i a b l e)$ $I = \int_{0}^{\frac{π}{2}} \frac{{(c o t x)}^{n}}{1 + {(c o t x)}^{n}} d x (a d d t h e t w o i n t e g r a l)$ $I + I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + {(c o t x)}^{n}} d x + \int_{0}^{\frac{π}{2}} \frac{{(c o t x)}^{n}}{1 + {(c o t x)}^{n}} d x$ $2 I = \int_{0}^{\frac{π}{2}} \frac{1 + {(c o t x)}^{n}}{1 + {(c o t x)}^{n}} d x \Leftrightarrow 2 I = \int_{0}^{\frac{π}{2}} 1 d x$ $2 I = x ∣_{0}^{\frac{π}{2}} + C$ $2 I = (\frac{π}{2} - 0)$ $I = \frac{π}{4}$
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