Advertisements
Advertisements
प्रश्न
Evaluate:
`int_(-1/2)^(1/2) cosx.log ((1+x)/(1-x))dx`
मूल्यांकन
उत्तर
Let, `int_(-1/2)^(1/2) cosx.log ((1+x)/(1-x))dx` ...(i) `["using property" int_-a^b f(x) = int_-a^b f(a+b-x) dx]`
`I=int_(-1/2)^(1/2) cos(-1/2+1/2-x).log [(1+(-1/2+1/2-x))/(1-(-1/2+1/2-x))]dx`
`= int_(-1/2)^(1/2) cos(-x).log ((1-x)/(1+x))dx`
`I=int_(-1/2)^(1/2) cos x log ((1-x)/(1+x)) dx` ...(ii)
On adding eq. (i) and eq (ii)
`I+I = int_(-1/2)^(1/2) cosxlog((1-x)/(1+x)) dx+cosxlog ((1-x)/(1+x))dx`
`2I = int_(-1/2)^(1/2) cosx[log((1+x)/(1-x))+log ((1-x)/(1+x))]dx` ...[using log m + log n = log mn]
`2I = int_(-1/2)^(1/2) cosxlog((1+x)/(1-x))((1-x)/(1+x))dx`
`2I = int_(-1/2)^(1/2) cosx log1 dx`
`2I = int_(-1/2)^(1/2) cos x (0)dx` ...[log 1 = 0]
2I = 0
I = 0
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
