i know the solution,,,,,,but i m trying to integrate it by other type,,,,problem is that my answer is not matching,,,,so can any one tell me where i m making the mistake,,,,here is my solution....

first just convert in to sinx and cosx,,,,so we getcosx-sinx/sinx+cosxdx  =  log[sinx+cosx]+c,,,,

now see this,,,log[{sinx/+cosx/}]

now we can make this as,,,,log[{cos(-x)}]  = log[{sin(-())}]+c

so finally we get....

but the answer is

chinmay...both ur answers are right..

1st consider..

answer a :log[sin(pi/4 +x)] + c

answer b: log[rt2sin(pi/4+x)] + d {i am using 'd' to indicate tht it is  a diff constant}

answer b can be written as log(rt2)+log[sin(pi/4 +x)] +d as logab=loga+logb

thts all.. the 'c' u got in the 1st case is equal to 'd' + log(rt2)..thts the only diff!!!

the mistake u have made is tht u have considered c,d as the same which is not true in this case

cheers