1/(1+sin(x)+cos(x))?

Question

plx help me i have no clue how to do this :?

Answer 1

If your goal is to simplify this, try and think of this as a binomial. Note that your question is equivalent to

1/[ ( sin(x) + cos(x) ) + 1 ]

Now, multiply top and bottom by the bottom's conjugate, which is (sin(x) + cos(x)) - 1. This gives us a difference of squares in the denominator.

[sin(x) + cos(x) - 1] / [ (sin(x) + cos(x))^2 - 1 ]

Expand the bottom

[sin(x) + cos(x) - 1] / [ sin^2(x) + 2sin(x)cos(x) + cos^2(x) - 1]

Notice the denominator, which has sin^2(x) + cos^2(x); this is equal to 1.

[sin(x) + cos(x) - 1] / [2sin(x)cos(x) + 1 - 1]

Now, we can simplify the bottom by eliminating the 1 and -1.

[sin(x) + cos(x) - 1] / [2sin(x)cos(x)]

Now, we can split this up into three fractions:

sin(x)/[2sin(x)cos(x)] + cos(x)/[2sin(x)cos(x)] - 1/[2sin(x)cos(x)]

Notice the first two fractions have cancellations. Cancelling appropriately,

1/[2cos(x)] + 1/[2sin(x)] - 1/[2sin(x)cos(x)]

Factor out the constants (which is 1/2 in this case):

(1/2) [1/cos(x)] + (1/2)[1/sin(x)] + (1/2)[1/[sin(x)cos(x)])

I'll leave you to figure out the details of the rest, but continuing on, we get

(1/2)sec(x) + (1/2)csc(x) + (1/2) [csc(x)sec(x)]

Answer 2

What are you supposed to do? Simplify it or integrate it?

Answer 3

To do what? What is it that you need to do?