Question Number 102790 by bobhans last updated on 11/Jul/20

I=2∫_0 ^(1/(√2))  ((sin^(−1) (x))/x) dx −∫_0 ^1  ((tan^(−1) (x))/x) dx

Answered by bramlex last updated on 11/Jul/20

set x = sin h in the first integral  and x = tan h in the second  integral yields   I=2∫_0 ^(π/4) ((h cos h dh)/(sin h)) −∫_0 ^(π/4)  ((h sec ^2 h dh)/(tan h))  I= ∫_0 ^(π/4) h cot h(2−sec ^2 h)dh  =∫_0 ^(π/4) h cot h(1−tan ^2 h) dh   =∫_0 ^(π/4) h(cot h−tan h)dh   by parts, where u = h , dv = (cot h−tan h)dh  I=h(ln(sin h)+ln(cos h))∣_0 ^(π/4) −∫_0 ^(π/4) ln(sin h)+ln(cos h)dh  I= h ln((1/2)sin (2h))∣^(π/4) _( 0) −∫_0 ^(π/4) ln((1/2)sin (2h))dh  I=−(π/4)ln(2)−∫_0 ^(π/4) (−ln2 +ln(sin (2h)))dh  I= −∫_0 ^(π/4) ln(sin (2h))dh  I=−(1/2)∫_0 ^(π/2) ln(sin t) dt ; via t = 2h  I= −(1/2). (−(π/2)ln(2))= ((π ln(2))/4) .