Evaluate ∫

1
−1
cot−1
1
√
1−x2
(cot−1
x
√
1−(x2)|x|
)∫
1
−1
cot−1
1
√
1−x2
(cot−1
x
√
1−(x2)|x|
)
Using ∫
b
a
f(x)dx=∫
b
a
f(a+b−x)dx
, I got:
2I=2π∫
1
0
cot−1(
1
√
1−x2
)dx
Then, letting x=sinθ
I=2∫
π/2
0
arctan(cosθ)cosθdθ
After this I tried integration by parts but it gets really complicated with that? How do I continue?

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