Question Number 53472 by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
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Commented by malwaan last updated on 23/Jan/19
$$\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{sir}\:? \\ $$$$\mathrm{please} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19
![I=∫_(−a) ^a ((cos^b x+1)/(c^x +1))dx a=12345678910π multiple of 2π b=123456789 odd number c=201620172018 even number now let t=−x dx=−dt ∫_a ^(−a) ((cos^b (−t)+1)/(c^(−t) +1))(−dt) ∫_a ^(−a) ((cos^b t+1)/((1/c^t )+1))×−dt ∫_(−a) ^a ((c^t (cos^b t+1))/(1+c^t ))dt now ∫_(−a) ^a f(x)dx=∫_(−a) ^a f(t)dt 2I=∫_(−a) ^a ((cos^b x+1)/(c^x +1))dx+∫_(−a) ^a ((c^t (cos^b t+1))/(1+c^t ))dt 2I=∫_(−a) ^a (1+cos^b t)dt I=(1/2)∫_(−a) ^a dt+(1/2)∫_(−a) ^a cos^b tdt I=(1/2)×2a+I_1 I=a+I_1 [(1/2)∫_(−a) ^a cos^b tdt=0] so answdr is I=a=12345678910π](Q53549.png)
$${I}=\int_{−{a}} ^{{a}} \frac{{cos}^{{b}} {x}+\mathrm{1}}{{c}^{{x}} +\mathrm{1}}{dx} \\ $$$${a}=\mathrm{12345678910}\pi\:\:{multiple}\:{of}\:\mathrm{2}\pi \\ $$$${b}=\mathrm{123456789}\:\:{odd}\:{number} \\ $$$${c}=\mathrm{201620172018}\:{even}\:{number} \\ $$$${now}\:{let}\:{t}=−{x}\:\:\:{dx}=−{dt} \\ $$$$\int_{{a}} ^{−{a}} \frac{{cos}^{{b}} \left(−{t}\right)+\mathrm{1}}{{c}^{−{t}} +\mathrm{1}}\left(−{dt}\right) \\ $$$$\int_{{a}} ^{−{a}} \frac{{cos}^{{b}} {t}+\mathrm{1}}{\frac{\mathrm{1}}{{c}^{{t}} }+\mathrm{1}}×−{dt} \\ $$$$\int_{−{a}} ^{{a}} \frac{{c}^{{t}} \left({cos}^{{b}} {t}+\mathrm{1}\right)}{\mathrm{1}+{c}^{{t}} }{dt} \\ $$$${now}\:\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\int_{−{a}} ^{{a}} {f}\left({t}\right){dt} \\ $$$$\mathrm{2}{I}=\int_{−{a}} ^{{a}} \frac{{cos}^{{b}} {x}+\mathrm{1}}{{c}^{{x}} +\mathrm{1}}{dx}+\int_{−{a}} ^{{a}} \frac{{c}^{{t}} \left({cos}^{{b}} {t}+\mathrm{1}\right)}{\mathrm{1}+{c}^{{t}} }{dt} \\ $$$$\mathrm{2}{I}=\int_{−{a}} ^{{a}} \left(\mathrm{1}+{cos}^{{b}} {t}\right){dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−{a}} ^{{a}} {dt}+\frac{\mathrm{1}}{\mathrm{2}}\int_{−{a}} ^{{a}} {cos}^{{b}} {tdt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{a}+{I}_{\mathrm{1}} \\ $$$${I}={a}+{I}_{\mathrm{1}} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{−{a}} ^{{a}} {cos}^{{b}} {tdt}=\mathrm{0}\right] \\ $$$${so}\:{answdr}\:{is}\:{I}={a}=\mathrm{12345678910}\pi \\ $$
Commented by malwaan last updated on 24/Jan/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\: \\ $$