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Question Number 65805 by ~ À ® @ 237 ~ last updated on 04/Aug/19
Prove that I_n =∫_0 ^(π/2) (dt/(1+(tant)^n )) does not depend of the term n deduces that ∫_0 ^∞ (dx/((x^(2035) +1)(x^2 +1)))=(π/4)
$$\:\:{Prove}\:{that}\:\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}+\left({tant}\right)^{{n}} }\:\:{does}\:{not}\:{depend}\:{of}\:{the}\:{term}\:{n} \\ $$$${deduces}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2035}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\pi}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 04/Aug/19
I_n =∫_0 ^(π/2) (dt/(1+(tant)^n )) changement t =(π/2)−x give I_n =−∫_0 ^(π/2) ((−dx)/(1+(1/(tan^n x)))) =∫_0 ^(π/2) ((tan^n x)/(1+tan^n x))dx =∫_0 ^(π/2) ((1+tan^n x−1)/(1+tan^n x))dx =(π/2) −I_n ⇒ 2I_n =(π/2) ⇒ I_n =(π/4) ∀n changement tant =x give I_n =∫_0 ^∞ (dx/((1+x^2 )(1+x^n ))) ⇒ ∫_0 ^∞ (dx/((1+x^2 )(1+x^n ))) =(π/4) let take n=2035 ⇒ ∫_0 ^∞ (dx/((x^(2035) +1)(1+x^2 ))) =(π/4)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}+\left({tant}\right)^{{n}} }\:\:{changement}\:{t}\:=\frac{\pi}{\mathrm{2}}−{x}\:{give} \\ $$$${I}_{{n}} =−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{−{dx}}{\mathrm{1}+\frac{\mathrm{1}}{{tan}^{{n}} {x}}}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tan}^{{n}} {x}}{\mathrm{1}+{tan}^{{n}} {x}}{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{{n}} {x}−\mathrm{1}}{\mathrm{1}+{tan}^{{n}} {x}}{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−{I}_{{n}} \:\Rightarrow\:\mathrm{2}{I}_{{n}} =\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}_{{n}} =\frac{\pi}{\mathrm{4}}\:\:\forall{n} \\ $$$${changement}\:{tant}\:={x}\:{give}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{{n}} \right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{{n}} \right)}\:=\frac{\pi}{\mathrm{4}}\:\:{let}\:{take}\:{n}=\mathrm{2035}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2035}} +\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\frac{\pi}{\mathrm{4}} \\ $$
Answered by Tanmay chaudhury last updated on 04/Aug/19
I(n)=∫_0 ^(π/2) ((cos^n t)/(sin^n t+cos^n t))dt.....(1) =∫_0 ^(π/2) ((cos^n ((π/2)−t))/(sin^n ((π/2)−t)+cos^n ((π/2)−t)))dt =∫_0 ^(π/2) ((sin^n t)/(cos^n t+sin^n t))dt.....(2) 2I(n)=∫_0 ^(π/2) dt.....(by adding (1) and 2) I(n)=(1/2)×(π/2)=(π/4)
$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{{n}} {t}}{{sin}^{{n}} {t}+{cos}^{{n}} {t}}{dt}…..\left(\mathrm{1}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{{n}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)}{{sin}^{{n}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)+{cos}^{{n}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{{n}} {t}}{{cos}^{{n}} {t}+{sin}^{{n}} {t}}{dt}…..\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{I}\left({n}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dt}…..\left({by}\:{adding}\:\left(\mathrm{1}\right)\:{and}\:\mathrm{2}\right) \\ $$$${I}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$
Answered by Tanmay chaudhury last updated on 04/Aug/19
x=tana dx=sec^2 ada ∫_0 ^(π/2) ((sec^2 ada)/(tan^(2035) a+1))×(1/(sec^2 a))da =(π/4)
$${x}={tana}\:\:\:{dx}={sec}^{\mathrm{2}} {ada} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {ada}}{{tan}^{\mathrm{2035}} {a}+\mathrm{1}}×\frac{\mathrm{1}}{{sec}^{\mathrm{2}} {a}}{da} \\ $$$$=\frac{\pi}{\mathrm{4}} \\ $$

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