CBSE Class 12-science Answered
evaluate integration of log(sinx) dx
Asked by jprathamesh023 | 23 Mar, 2024, 16:01: PM
Let us use the following property of definie integration ,
Then eqn(1) becomes
By adding (1) and (2) , we get
Let us use the substitution u = 2x ; du = 2 dx ;
when x = 0 , u = 0 ; when x = π/2 , u = π ;
Now let us use the following property of definite integration
Using the above property integration I1 is
above integration is written because log(sin(π-u)) = log(sinu)
RHS of above expression is same as I as given in eqn.(1).
Hence I1 = I
Then from eqn.(3) , we get
I = - log2 ( π/2)
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Answered by Thiyagarajan K | 23 Mar, 2024, 21:21: PM
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