plz give sum hint!!!

1. ∫cos2x/ (1+tanx) dx
2. ∫[(x/e)x+(e/x)x] lnx dx

6 Answers

1357
Manish Shankar ·

1.
cos2x/(1+tanx)=sec2x/[sec4x(1+tanx)]=sec2x/[(1+tan2x)2(1+tanx)

I don't have any better approach

1357
Manish Shankar ·

2.
[(x/e)x+(e/x)x]lnx=(x/e)xlnx+(e/x)xlnx
taking first one
(x/e)xlnx dx

taking (x/e)x=t

then (x/e)xlnx dx=dt

∫(x/e)xlnx dx=∫dt=t+C=(x/e)x+C

Similarly fond for other

11
Devil ·

2nd one can be esily killed by parts.

341
Hari Shankar ·

http://www.goiit.com/posts/list/integration-integration-plz-can-anyone-help-me-77629.htm#379975

1
Surbhi Agrawal ·

can u explain the first one!.. i got the 2nd one!!

11
Mani Pal Singh ·

In the 1st u have to know that d/dx tanx =sec2x
so

multiply num and denom by sec4x

upper sec2x and neeche sec4x+tanxsec4x aa jaega

then put tanx=t and sec2xdx=dt

neeche sec4x ko write in terms of tan x and substitute to get ur answer

Your Answer

Close [X]