Menu Close

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  ∫_( 0) ^π  f (cos x) dx = 2∫_( 0) ^(π/2)  f (cos x) dx
$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function},\:\mathrm{then} \\ $$$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:{f}\:\left(\mathrm{cos}\:{x}\right)\:{dx}\:=\:\mathrm{2}\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:{f}\:\left(\mathrm{cos}\:{x}\right)\:{dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
re check the question...
$${re}\:{check}\:{the}\:{question}… \\ $$
Commented by maxmathsup by imad last updated on 15/Jan/19
∫_0 ^π f(cosx)dx =∫_0 ^(π/2) f(cosx)dx +∫_(π/2) ^π f(cosx)dx but  ∫_(π/2) ^π f(cosx)dx =_(x =(π/2)+t)   ∫_0 ^(π/2) f(−sint)dt =−∫_0 ^(π/2) f(sint)dt (f is odd)  =−∫_0 ^(π/2) f(cos((π/(2 ))−t))dt =_((π/(2  ))−t =u)   −∫_(π/2) ^0 f(cosu)(−du)  =−∫_0 ^(π/2) f(cosu)du ⇒∫_0 ^π f(cosx)dx =0   if f is even we get ∫_(π/2) ^π  f(cost)dt =_(x=(π/2)+t)    ∫_0 ^(π/2) f(−sint)dt =∫_0 ^(π/2) f(sint)dt  =∫_0 ^(π/2) f(cos((π/2)−t))dt =_((π/2)−t=u)   − ∫_(π/2) ^0  f(cosu)du=∫_0 ^(π/2) f(cosu)du ⇒  ∫_0 ^π f(cosx)dx =2 ∫_0 ^(π/2) f(cosx)dx so there is a error in the question!...
$$\int_{\mathrm{0}} ^{\pi} {f}\left({cosx}\right){dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cosx}\right){dx}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {f}\left({cosx}\right){dx}\:{but} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {f}\left({cosx}\right){dx}\:=_{{x}\:=\frac{\pi}{\mathrm{2}}+{t}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left(−{sint}\right){dt}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({sint}\right){dt}\:\left({f}\:{is}\:{odd}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cos}\left(\frac{\pi}{\mathrm{2}\:}−{t}\right)\right){dt}\:=_{\frac{\pi}{\mathrm{2}\:\:}−{t}\:={u}} \:\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {f}\left({cosu}\right)\left(−{du}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cosu}\right){du}\:\Rightarrow\int_{\mathrm{0}} ^{\pi} {f}\left({cosx}\right){dx}\:=\mathrm{0}\: \\ $$$${if}\:{f}\:{is}\:{even}\:{we}\:{get}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{f}\left({cost}\right){dt}\:=_{{x}=\frac{\pi}{\mathrm{2}}+{t}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left(−{sint}\right){dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({sint}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cos}\left(\frac{\pi}{\mathrm{2}}−{t}\right)\right){dt}\:=_{\frac{\pi}{\mathrm{2}}−{t}={u}} \:\:−\:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \:{f}\left({cosu}\right){du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cosu}\right){du}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} {f}\left({cosx}\right){dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({cosx}\right){dx}\:{so}\:{there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}!… \\ $$
Commented by gunawan last updated on 16/Jan/19
thank you Sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$
Commented by maxmathsup by imad last updated on 16/Jan/19
you are welcome
$${you}\:{are}\:{welcome} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *