Question-42593

Question Number 42593 by aseerimad last updated on 28/Aug/18

Commented by maxmathsup by imad last updated on 28/Aug/18

w e h a v e I = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{4} x + {s i n}^{4} x - {s i n}^{4} x}{{s i n}^{4} x + {c o s}^{4} x} d x = \frac{π}{2} - \int_{0}^{\frac{π}{2}} \frac{{s i n}^{4} x}{{s i n}^{4} x + {c o s}^{4} x} d x b u t

c h a n g e m e n t x = \frac{π}{2} - t g i v e \int_{0}^{\frac{π}{2}} \frac{{s i n}^{4} x}{{s i n}^{4} x + {c o s}^{4} x} d x = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{4} t}{{c o s}^{4} x + {s i n}^{4} x} d t

= I \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} .

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

I = \int_{0}^{\frac{Π}{2}} f (x) d x = \int_{0}^{\frac{Π}{4}} f (\frac{Π}{2} - x) d x

I = \int_{0}^{\frac{Π}{2}} \frac{{c o s}^{4} x}{{s i n}^{4} x + {c o s}^{4} x} d x = \int_{0}^{\frac{Π}{4}} \frac{{c o s}^{4} (\frac{Π}{2} - x)}{{s i n}^{4} (\frac{Π}{2} - x) + {c o s}^{4} (\frac{Π}{2} - x)} d x

I = \int_{0}^{\frac{Π}{2}} \frac{{s i n}^{4} x}{{c o s}^{4} x + {s i n}^{4} x} d x

2 I = \int_{0}^{\frac{Π}{2}} \frac{{c o s}^{4} x + {s i n}^{4} x}{{c o s}^{4} x + {s i n}^{4} x} d x = \int_{0}^{\frac{Π}{2}} d x

2 I =∣ x ∣_{0}^{\frac{Π}{2}} = \frac{Π}{2}

I = \frac{Π}{4}