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Question Number 65805 by ~ À ® @ 237 ~ last updated on 04/Aug/19

$$Provethat{I}_n={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{\left(tant\right)}^n}doesnotdependofthetermn$$$$deducesthat$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{(x^{2035}+1)(x^2+1)}=\frac{\pi }{4}$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$I_n={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+{\left(tant\right)}^n}changementt=\frac{\pi }{2}-xgive$$$$I_n=-{\int }_0^{\frac{\pi }{2}}\frac{-dx}{1+\frac{1}{{tan}^{n}x}}={\int }_0^{\frac{\pi }{2}}\frac{{tan}^{n}x}{1+{tan}^{n}x}dx={\int }_0^{\frac{\pi }{2}}\frac{1+{tan}^{n}x-1}{1+{tan}^{n}x}dx$$$$=\frac{\pi }{2}-I_n\Rightarrow 2I_n=\frac{\pi }{2}\Rightarrow I_n=\frac{\pi }{4}\mathrm{\forall }n$$$$changementtant=xgive{I}_n={\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x^n)}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(1+x^n)}=\frac{\pi }{4}lettaken=2035\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{(x^{2035}+1)(1+x^2)}=\frac{\pi }{4}$$

Answered by Tanmay chaudhury last updated on 04/Aug/19

$$I\left(n\right)={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{n}t}{{sin}^{n}t+{cos}^{n}t}dt.....\left(1\right)$$$$={\int }_0^{\frac{\pi }{2}}\frac{{cos}^n(\frac{\pi }{2}-t)}{{sin}^n(\frac{\pi }{2}-t)+{cos}^n(\frac{\pi }{2}-t)}dt$$$$={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{n}t}{{cos}^{n}t+{sin}^{n}t}dt.....\left(2\right)$$$$2I\left(n\right)={\int }_0^{\frac{\pi }{2}}dt.....\left(byadding\left(1\right)and2\right)$$$$I\left(n\right)=\frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{4}$$

Answered by Tanmay chaudhury last updated on 04/Aug/19

$$x=tanadx={sec}^{2}ada$$$${\int }_0^{\frac{\pi }{2}}\frac{{sec}^{2}ada}{{tan}^{2035}a+1}\times \frac{1}{{sec}^{2}a}da$$$$=\frac{\pi }{4}$$

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