Prove the following integral
I=π2∫0dx1+sin2(tanx)=π2√2(e2+3−2√2e2−3+2√2)This integral result was calculated using Mathematica and
I like this integral. But I can’t solve it.My idea:
Let tanx=t⟹dx=11+t2dt
so
I=∞∫0dt1+sin2t⋅11+t2then 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=2∫∞01(3−cos2t)(1+t2) dt
Then using the identity ∞∑k=0akcos(kx)=1−acosx1−2acosx+a2, |a|<1
we have
1+2∞∑k=1akcos(kx)=1+2(1−acosx1−2acosx+a2−1)=1−a21−2acosx+a2
Therefore,
∫∞01(1−2acos2x+a2)(1+x2) dx=11−a2∫∞0(1+2∞∑k=1akcos(2kx)) 11+x2 dx=π211−a2+21−a2∞∑k=1ak∫∞0cos(2kx)1+x2 dx=π211−a2+π1−a2∞∑k=1(ae2)k=π211−a2+π1−a2a/e21−a/e2=π211−a2(1+2ae2−a)=π211−a2e2+ae2−a
Now rewrite the integral as
11+a2∫∞01(1−2a1+a2cos2x)(1+x2) dx
and let a=3−2√2.
Then
11+a2∫∞01(1−13cos2x)(1+x2) dx=31+a2∫∞01(3−cos2x)(1+x2) dx=π211−a2e2+3−2√2e2−3+2√2
which implies
∫π/2011+sin2(tanx) dx=π31+a21−a2e2+3−2√2e2−3+2√2=π332√2e2+3−2√2e2−3+2√2=π2√2e2+3−2√2e2−3+2√2
Attribution
Source : Link , Question Author : math110 , Answer Author : Random Variable