Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 201515 by sonukgindia last updated on 08/Dec/23

	Answered by Calculusboy last updated on 08/Dec/23

	$S o l u t i o n : I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + {(\frac{1}{t a n x})}^{\sqrt{2}}} d x \Leftrightarrow I = \int_{0}^{\frac{π}{2}} \frac{1}{\frac{{(t a n x)}^{\sqrt{2}} + 1}{{(t a n x)}^{\sqrt{2}}}} d x$ $I = \int_{0}^{\frac{π}{2}} \frac{{(t a n x)}^{\sqrt{2}}}{1 + {(t a n x)}^{\sqrt{2}}} d x l e t y = \frac{π}{2} - x d y = - d x$ $w h e n x = \frac{π}{2} y = 0 a n d w h e n x = 0 y = \frac{π}{2}$ $I = \int_{\frac{π}{2}}^{0} \frac{{[t a n (\frac{π}{2} - y)]}^{\sqrt{2}}}{1 + {[t a n (\frac{π}{2} - y)]}^{\sqrt{2}}} (- d y) \Leftrightarrow I = \int_{0}^{\frac{π}{2}} \frac{{(c o t y)}^{\sqrt{2}}}{1 + {(c o t y)}^{\sqrt{2}}} d y$ $N b : t a n (\frac{π}{2} - y) = c o t y a n d c h a n g i n g o f v a r i a b l e$ $I = \int_{0}^{\frac{π}{2}} \frac{{(c o t x)}^{\sqrt{2}}}{1 + {(c o t x)}^{\sqrt{2}}} d x (a d d t h e t w o i n t e g r a l)$ $I + I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + {(c o t x)}^{\sqrt{2}}} d x + \int_{0}^{\frac{π}{2}} \frac{{(c o t x)}^{\sqrt{2}}}{1 + {(c o t x)}^{\sqrt{2}}} d x$ $2 I = \int_{0}^{\frac{π}{2}} \frac{1 + {(c o t x)}^{\sqrt{2}}}{1 + {(c o t x)}^{\sqrt{2}}} d x \Leftrightarrow 2 I = \int_{0}^{\frac{π}{2}} 1 d x$ $2 I = x ∣_{0}^{\frac{π}{2}} + C$ $2 I = (\frac{π}{2} - 0)$ $I = \frac{π}{4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum