0-pi-2-f-x-f-x-f-pi-2-x-dx-

Question Number 59976 by soufiane last updated on 16/May/19

\underset{0}{\int^{π / 2}} \frac{f (x)}{f (x) + f (\frac{π}{2} - x)} d x =

Answered by tanmay last updated on 16/May/19

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

t = \frac{π}{2} - x

d t = - d x

I = \int_{\frac{π}{2}}^{0} \frac{f (\frac{π}{2} - t)}{f (\frac{π}{2} - t) + f (t)} \times - d t

I = \int_{0}^{\frac{π}{2}} \frac{f (\frac{π}{2} - t)}{f (\frac{π}{2} - t) + f (t)} d t \to \int_{0}^{\frac{π}{2}} \frac{f (\frac{π}{2} - x)}{f (\frac{π}{2} - x) + f (x)} d x

n o w \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t

s o

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

2 I = \int_{0}^{\frac{π}{2}} d x

2 I = \frac{π}{2} \to I = \frac{π}{4}

d i r e c t f o r m u l a

\int_{a}^{b} f (x) d x = \int_{0}^{b} f (a + b - x) d x

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

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

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

2 I = \int_{0}^{\frac{π}{2}} d x

I = \frac{π}{4}

Answered by Prithwish sen last updated on 16/May/19

I = \int_{0}^{\frac{π}{2}} \frac{f (\frac{π}{2} - x)}{f (\frac{π}{2} - x) + f (x)} dx

∴ 2 I = \int_{0}^{\frac{π}{2}} \frac{f (\frac{π}{2} - x) + f (x)}{f (\frac{π}{2} - x) + f (x)} dx

= \int_{0}^{\frac{π}{2}} dx = \frac{π}{2}

∴ I = \frac{π}{4}