Prove π/2∫0dx1+sin2(tanx)=π2√2(e2+3−2√2e2−3+2√2)\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)

Prove the following integral
I=π20dx1+sin2(tanx)=π22(e2+322e23+22)

This integral result was calculated using Mathematica and
I like this integral. But I can’t solve it.

My idea:

Let tanx=tdx=11+t2dt
so
I=0dt1+sin2t11+t2

then I can’t proceed. Can you help me? Thank you.

Answer

First note that

π/2011+sin2(tanx) dx=01(1+sin2t)(1+t2) dt=201(3cos2t)(1+t2) dt

Then using the identity k=0akcos(kx)=1acosx12acosx+a2,  |a|<1

we have

1+2k=1akcos(kx)=1+2(1acosx12acosx+a21)=1a212acosx+a2

Therefore,

01(12acos2x+a2)(1+x2) dx=11a20(1+2k=1akcos(2kx)) 11+x2 dx=π211a2+21a2k=1ak0cos(2kx)1+x2 dx=π211a2+π1a2k=1(ae2)k=π211a2+π1a2a/e21a/e2=π211a2(1+2ae2a)=π211a2e2+ae2a

Now rewrite the integral as

11+a201(12a1+a2cos2x)(1+x2) dx

and let a=322.

Then

11+a201(113cos2x)(1+x2) dx=31+a201(3cos2x)(1+x2) dx=π211a2e2+322e23+22

which implies

π/2011+sin2(tanx) dx=π31+a21a2e2+322e23+22=π3322e2+322e23+22=π22e2+322e23+22

Attribution
Source : Link , Question Author : math110 , Answer Author : Random Variable

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