IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

DON007

hey toshi look.............

_{I = [0 ]}^{[pie/2 ]} log tanx dx ...........(1)

_{I= [0 ]}^{[pie/2 ]} log tan(pie/2- x )dx

(using the formula...........

_{[0 ]}^{[a ]} f(x)dx = _{[0 ]}^{[a ]} f(a-x) dx = 0

I= _{[0 ]}^{[pie/2 ]} log cotx dx .....(3)

add 1 and 3 we get.............

2I =
_{[0 ]}^{[pie/2 ]} (log tanx + log cot x) dx

_{[0 ]}^{[pie/2 ]} log (tanx*cotx) dx

_{2I=[0 ]}^{[pie/2 ]} log 1 dx

_{2I= [0 ]}^{[pie/2 ]} 0 dx

2I=0...................ANSWER

HOPE U UNDERSTAND.......................

THANKU............