yar integration sikhado Yar ki toh
kardo iska Integration bhaiyo
1)1/sinx+cosx+1
Answers
Well, if you only have:
1 + Sin[x] + Cos[x]
-------------------
1 - Sin[x] + Cos[x]
and are told to simplify, here's what to try:
First, I performed long division on it, so I got:
2 Sin[x]
1 + -------------------
1 - Sin[x] + Cos[x]
But then I didn't know what to do with the remainder. So I started
over, and recognized that you could group the terms like this:
(1+Cos[x]) + Sin[x]
-------------------
(1+Cos[x]) - Sin[x].
So the numerator is of a form like A+B, and the denominator is like
A-B. We multiply the numerator and denominator by A+B, so that the
denominator becomes a difference of squares, A^2 - B^2. We get:
((1+Cos[x]) + Sin[x])^2
-----------------------
(1+Cos[x])^2 - Sin[x]^2
(1+Cos[x])^2 + 2(1+Cos[x])Sin[x] + Sin[x]^2
= -------------------------------------------
1 + 2 Cos[x] + Cos[x]^2 - Sin[x]^2
which is kind of messy, until you notice the trig identity:
1 - Sin[x]^2 = Cos[x]^2
and complete expanding the numerator. This gives:
1 + 2 Cos[x] + Cos[x]^2 + 2 Sin[x] + 2 Sin[x]Cos[x] + Sin[x]^2
--------------------------------------------------------------
2 Cos[x] + 2 Cos[x]^2
2 + 2(Cos[x] + Sin[x] + Sin[x]Cos[x])
= -------------------------------------
2 Cos[x](1 + Cos[x])
and when we cancel out the 2's, we get:
1 + Cos[x] + Sin[x] + Sin[x]Cos[x]
---------------------------------- .
Cos[x](1 + Cos[x])
But notice the numerator can be factored into (1+Sin[x])(1+Cos[x]),
and so we can cancel out a common factor of 1+Cos[x] to give the final
result:
1 + Sin[x]
---------- .
Cos[x]