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Integral Calculus
IIT BOMBAY OLYMPIAD QUESTION INTEGRATION SOLVE THIS !!integral-((pi/2)(0)[log(tanx)])
I= int ( pi/2 , 0 ) log(tan(x)) dx
I= int (pi/2 , 0) log(cos(x)) dx
Adding above
2I= int (pi/2 , 0) log (1) dx
I= 0
thank u
are you sure ???
Coz doesnt definite integral gives area under curve here we have logtanx tanx-->inf. to pi/2 the are i can t think can be zero .
yeah ... so probably below the point x = pi/4 where y = 0 ..... area is negative and its magnitude equals d area of d part above d x-axis .....
amazing....
not the graph...ur efforts to help ...
graph toh koi bhi daal sakta hai ....net se ...calculators se ...aur ...etc etc....
The Area cannot be zero.
first integrate from 0 to pi/4 and then multiply by 2,not considering the sign.
That is, 0.91596559`*2 = 0.915966,which is the required answer.
A mistake-the result is
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I= int ( pi/2 , 0 ) log(tan(x)) dx
I= int (pi/2 , 0) log(cos(x)) dx
Adding above
2I= int (pi/2 , 0) log (1) dx
I= 0
thank u