Question Number 114808 by mnjuly1970 last updated on 21/Sep/20

      ...nice math...         evaluate ::                           I =∫_0 ^( (π/4)) ((ln(1+sin(x)))/(cos(x)))dx=???     ...m.n.july.1970...

Answered by mathdave last updated on 22/Sep/20

solution  let  I=∫_0 ^(π/4) ((ln(1+sinx))/(cosx))dx=∫_0 ^(π/4) ((ln(1+sinx))/(1−sin^2 x))cosxdx  put y=sinx  I=∫_0 ^(1/( (√2))) ((ln(1+y))/(1−y^2 ))dy=(1/2)(∫_0 ^(1/( (√2))) ((ln(1+y))/(1−y))dy+∫_0 ^(1/( (√2))) ((ln(1+y))/(1+y))dy)=(1/2)(A+B)  put y=1−2x in  A  let  A=∫_((1/2)−((√2)/4)) ^(1/( (√2))) ((ln(2−2x))/x)dx=∫_((1/2)−((√2)/4)) ^(1/2) (((ln(1−x))/x)+((ln2)/x))dx  A=∫_((1/2)−((√2)/4)) ^(1/2) ((ln(1−x))/x)dx+ln2∫_((1/2)−((√2)/4)) ^(1/2) (1/x)dx  A=[−Li_2 (x)]_((1/2)−((√2)/4)) ^(1/2) +ln2[lnx]_((1/2)−((√2)/4)) ^(1/2)   A=−Li_2 ((1/2))+Li_2 ((1/2)−((√2)/4))+ln2[ln((1/2))−ln((1/2)−((√2)/4))]  A=−ln^2 (2)−Li_2 ((1/2))+Li_2 ((1/2)−((√2)/4))−ln2ln(2−(√2))+2ln^2 (2)  A=(3/2)ln^2 (2)−(π^2 /(12))+Li_2 ((1/2)−((√2)/4))−ln2ln(2−(√2)).....(1) and  B=∫_0 ^(1/( (√2))) ((ln(1+y))/(1+y))dy   using IBP  B=ln^2 (1+y)∣_0 ^(1/( (√2))) −∫_0 ^(1/( (√2))) ((ln(1+y))/(1+y))dy  B=(1/2)ln^2 (1+(1/( (√2))))=(1/2)ln^2 (((2+(√2))/2))=(1/2)[ln(((2+(√2))/2))]^2   B=(1/2)[ln(2+(√2))−ln2]^2   B=(1/2)ln^2 (2+(√2))+(1/2)ln^2 (2)−ln2ln(2+(√2))....(2)  but I=(1/2)(A+B)  I=(1/2)((3/2)ln^2 (2)−(π^2 /(12))+Li_2 ((1/2)−((√2)/4))−ln2ln(2−(√2))+(1/2)ln^2 (2+(√2))−ln2ln(2+(√2))+(1/2)ln^2 (2))  I=(1/2)(2ln^2 (2)−(π^2 /(12))+Li_2 ((1/2)−((√2)/4))−ln2[ln(2−(√2))+ln(2+(√2))]+(1/2)ln^2 (2+(√2)))  I=(1/2)(2ln^2 (2)−(π^2 /(12))+Li_2 ((1/2)−((√2)/4))−ln^2 (2)+(1/2)ln^2 (2+(√2)))  I=(1/2)Li_2 ((1/2)−((√2)/4))−(π^2 /(24))+(1/4)ln^2 (2+(√2))+(1/2)ln^2 (2)  ∵∫_0 ^(π/4) ((ln(1+sinx))/(cosx))dx=(1/2)Li_2 ((1/2)−((√2)/4))−(π^2 /(24))+(1/4)ln^2 (2+(√2))+(1/2)ln^2 (2)  by mathdave(22/09/2020)

Commented bymnjuly1970 last updated on 22/Sep/20

thank you so much mr dave.very  nice ...

Commented byTawa11 last updated on 06/Sep/21

great sir