Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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

$${you}\:{are}\:{welcome} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com