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Integral Calculus
Comments (3)
when sinx is odd multiples of pie/2 then in some cases the value is -1, then in the denominator 0 appears and the functiom is not determinable.
As, only for continous function there is indefinite integration,i think it is not integrable.Once again you check the question.
If, i am wrong i am sorry.
tell me whether i am correct or not?
reply is must.
Let I = 
substitute t = pi/2 - x
u get
I = 
= 
= 
= - 2 . 21/2 . sin (t/2) + 2 / 21/2 . log | tan ( pi/4 + t/4 ) | + C
Now, substitiute 't' back
Thus,
I = -2. root(2). sin ( pi/4 - x/2 ) + root(2). log | tan ( 3pi/8 - x/4 ) | + C
this is the final expression....
now u can easily find f(x) and g(x) by comparing with given expression.
ask if u dint get it !!
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