\int_0^{1398 \pi} (\sin^2 x)^{\cos^2 x}dx=?

Lucille Davidson

Lucille Davidson

Answered question

2022-01-17

01398π(sin2x)cos2xdx=?

Answer & Explanation

esfloravaou

esfloravaou

Beginner2022-01-18Added 43 answers

It's a bit more pleasant to use e.g. 01x1x+(1x)xx(1x)dx than
0π((sin2x)cos2x+(cos2x)sin2x)dx
Note:
If we define pn(x) as a polynomial of degree n , we get:
01x1x+(1x)xx(1x)dx=22k=0p2k(ln2)22k(2k+1)!
p0(x)=1
p2(x)=x2+2x1
p4(x)=x4+4x36x2+60x+65
p6=x6+6x515x4+300x3+975x2+3750x+6687
p8(x)=x8+8x728x6+4550x4+35000x3+187236x2+769272x+1305089

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