Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 105455 by john santu last updated on 29/Jul/20

∫_(−1) ^1 ((e^x −e^(−x) )/(cos x)) dx

$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{cos}\:{x}}\:{dx}\: \\ $$

Answered by 1549442205PVT last updated on 29/Jul/20

Put f(x)= ((e^x −e^(−x) )/(cos x))⇒f(−x)=((e^(−x) −e^x )/(cos(−x))) =−((e^x −e^(−x) )/(cosx))=−f(x)(as cos(−x)=cosx) which shows that f(x) is an odd function ⇒∫_(−a) ^a f(x)dx=∫_(−a) ^0 f(x)dx+∫_0 ^a f(x)dx = _(x=−t) ∫_a ^0 f(−t)(−1)dt+∫_0 ^a f(x)dx ∫_a ^0 (−f(t))(−1)dt+∫_0 ^a f(x)dx(as f(−t)=−f(t)) =∫_a ^0 f(t)dt+∫_0 ^a f(x)dx =−∫_0 ^a f(t)dt+∫_0 ^a f(x)dx=0

$$\mathrm{Put}\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{cos}\:{x}}\Rightarrow\mathrm{f}\left(−\mathrm{x}\right)=\frac{\mathrm{e}^{−\mathrm{x}} −\mathrm{e}^{\mathrm{x}} }{\mathrm{cos}\left(−\mathrm{x}\right)} \\ $$$$=−\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{cosx}}=−\mathrm{f}\left(\mathrm{x}\right)\left(\mathrm{as}\:\mathrm{cos}\left(−\mathrm{x}\right)=\mathrm{cosx}\right) \\ $$$$\mathrm{which}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function} \\ $$$$\Rightarrow\int_{−\mathrm{a}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{−\mathrm{a}} ^{\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\underset{\boldsymbol{\mathrm{x}}=−\boldsymbol{\mathrm{t}}} {=\:\:}\:\:\int_{\mathrm{a}} ^{\mathrm{0}} \mathrm{f}\left(−\mathrm{t}\right)\left(−\mathrm{1}\right)\mathrm{dt}+\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\int_{\mathrm{a}} ^{\mathrm{0}} \left(−\mathrm{f}\left(\mathrm{t}\right)\right)\left(−\mathrm{1}\right)\mathrm{dt}+\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\left(\mathrm{as}\:\mathrm{f}\left(−\mathrm{t}\right)=−\mathrm{f}\left(\mathrm{t}\right)\right) \\ $$$$=\int_{\mathrm{a}} ^{\mathrm{0}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}+\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}+\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{0} \\ $$

Answered by Dwaipayan Shikari last updated on 29/Jul/20

∫_(−1) ^1 ((e^x −e^(−x) )/(cosx))dx=∫_(−1) ^1 ((e^(−x) −e^x )/(cos(−x)))dx=I 2I=∫((e^x −e^(−x) +e^(−x) −e^x )/(cosx)) I=0

$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{e}^{{x}} −{e}^{−{x}} }{{cosx}}{dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{e}^{−{x}} −{e}^{{x}} }{{cos}\left(−{x}\right)}{dx}={I} \\ $$$$\mathrm{2}{I}=\int\frac{{e}^{{x}} −{e}^{−{x}} +{e}^{−{x}} −{e}^{{x}} }{{cosx}} \\ $$$${I}=\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 29/Jul/20

we have ((e^x −e^(−x) )/(cosx)) =2 ×((sh(x))/(cosx)) this fuction is odd ⇒ ∫_(−1) ^1 ((e^x −e^(−x) )/(cosx))dx =0

$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{cosx}}\:=\mathrm{2}\:×\frac{\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{cosx}}\:\:\mathrm{this}\:\mathrm{fuction}\:\mathrm{is}\:\mathrm{odd}\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{cosx}}\mathrm{dx}\:=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com