If-f-x-is-an-odd-function-then-0-pi-f-cos-x-dx-2-0-pi-2-f-cos-x-dx-

Question Number 52949 by gunawan last updated on 15/Jan/19

If f (x) is an odd function, then

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

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

r e c h e c k t h e q u e s t i o n \dots

Commented by maxmathsup by imad last updated on 15/Jan/19

\int_{0}^{π} f (c o s x) d x = \int_{0}^{\frac{π}{2}} f (c o s x) d x + \int_{\frac{π}{2}}^{π} f (c o s x) d x b u t

\int_{\frac{π}{2}}^{π} f (c o s x) d x =_{x = \frac{π}{2} + t} \int_{0}^{\frac{π}{2}} f (- s i n t) d t = - \int_{0}^{\frac{π}{2}} f (s i n t) d t (f i s o d d)

= - \int_{0}^{\frac{π}{2}} f (c o s (\frac{π}{2} - t)) d t =_{\frac{π}{2} - t = u} - \int_{\frac{π}{2}}^{0} f (c o s u) (- d u)

= - \int_{0}^{\frac{π}{2}} f (c o s u) d u \Rightarrow \int_{0}^{π} f (c o s x) d x = 0

i f f i s e v e n w e g e t \int_{\frac{π}{2}}^{π} f (c o s t) d t =_{x = \frac{π}{2} + t} \int_{0}^{\frac{π}{2}} f (- s i n t) d t = \int_{0}^{\frac{π}{2}} f (s i n t) d t

= \int_{0}^{\frac{π}{2}} f (c o s (\frac{π}{2} - t)) d t =_{\frac{π}{2} - t = u} - \int_{\frac{π}{2}}^{0} f (c o s u) d u = \int_{0}^{\frac{π}{2}} f (c o s u) d u \Rightarrow

\int_{0}^{π} f (c o s x) d x = 2 \int_{0}^{\frac{π}{2}} f (c o s x) d x s o t h e r e i s a e r r o r i n t h e q u e s t i o n! \dots

Commented by gunawan last updated on 16/Jan/19

thank you Sir

Commented by maxmathsup by imad last updated on 16/Jan/19

y o u a r e w e l c o m e