IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

tabassumayya

_{[ ]}

^{[ ]} log(1+tanx)

caprio.nups

solve it by 'uv' method..taking 1 as first function and log part as second function..

vinod

Whats the upper and lower limit plzz be clear !!!!

after knowing limits solve the sum by using the property of definate integrals

_{[ b]}

^{[ a]} Log[1+tan(x)] = _{[ b]}

^{[ a]} Log[1+tan((a+b)-x)]

neal

This question will be solved easily if sum of the limits is

/4

vinod

The answer will be

/8[log2] if the upper and the lower limit are

/4 and 0 respectively.

_{[ 0]}

^[

/4^] Log[1+tan(x)]dx = Y -------------(1)

_{[ 0]}

^[

/4^] Log[1+tan(

/4 - x)]dx = Y

_{[0 ]}

^[

/4^] Log[1+[1-tan(x)]/[1+tan(x)] ]dx= Y

_{[0 ]}

^[

/4^] Log[2/[1+tan(x)] ]dx= Y ------------(2)

Adding (1) and (2)., we get

_{[0 ]}

^[

/4^] log[2{(1+tan(x)}/{1+tan(x)} ]dx = 2Y

Log(2) _{[0 ]}

^[

/4^] dx = 2Y

Y= (

/4)(1/2)log(2)

Y = (

/8)log(2)

cheers !

aditya_arora04

Hi,

According to me, the limits should be provided. I don't think that it can be solved using indefinite integration

deepak_agarwal

hii..

well vinod is absolutely right...limits need to be specified...summaya can u provide us the limits...bbye

deepu_0088

i am not able to solve it but i know the ans