IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

toshi

prove that:

_{[0 ]}

^{[pie/2 ]} log tanx dx = 0

plzz show all the steps!
thanks!

Arjun

_{[0 ]}

^{[pie/2 ]} log(tanx).dx

_{[0 ]}

^{[ pie/2]} logsinx -_{[ 0]}

^{[ pie/2]} logcosx

_{[ 0]}

^{[ pie/2]} log sin(pi/2-x) - _{[0 ]}

^{[ pi/2]} logcosx

_{[ ]}

^{[ ]} log cosx -_{[ ]}

^{[ ]} log cosx

this is using propeties of defenite integrals then we get the ans

as

answer is 0

thank u.

plzz rate me!

vinod

Yep he is right!!!!!!

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............