What is the integreal of 1 / 1 -sec x?

Question

What is the integral of 1 / 1 -sec x (if possible by using the "Multiplication by form of 1 )

Answer 1

∫ ( 1/(1 - sec(x)) dx )

We can't solve this directly; let's use identities.

Multiply the top and bottom by (1 + sec(x)) gives us

(1 + sec(x)) / (1 - sec²(x))

The bottom is equal to tan²(x).

(1 + sec(x)) / tan²(x)

Splitting this up into two fractions,

1/tan²(x) + sec(x)/tan²(x)

cot²(x) + sec(x)/tan²(x)

Change the second fraction to sines and cosines, and reduce to a simple fraction.

cot²(x) + [ 1/cos(x) ] / [sin²(x) / cos²(x)]

cot²(x) + cos(x) / sin²(x)

Use the identity cot²(x) = csc²(x) - 1

csc²(x) - 1 + cos(x)/sin²(x)

And now, we can take the integral of this.

∫ (csc²(x) dx) - ∫ (1 dx) + ∫ (cos(x)/sin²(x) dx )

The first two integrals are easy; -csc²(x) is a known derivative, so it follows that csc²(x) will differ by (-1), and its integral will be cot(x).

cot(x) - x + ∫ (cos(x)/sin²(x) dx )

To solve this, we use substitution.

Let u = sin(x). Then
du = cos(x) dx.

cot(x) - x + ∫ ( 1/u² du)

cot(x) - x + ∫ (uˉ² du )

Using the reverse power rule,

cot(x) - x + (-1)uˉ¹ + C

cot(x) - x - (1/u) + C

But u = sin(x), so

cot(x) - x - 1/sin(x) + C

cot(x) - x - csc(x) + C

Answer 2

I assume you meant to group this with parens:
∫1/(1-sec(x)) dx =
∫(1+sec(x)) / (1-sec(²x)) dx =
∫(1+sec(x)) / (-tan²(x)) dx =
-∫(cot²(x) + cos(x)/sin²(x)) dx =
-∫(csc²(x)-1)dx + 1/sin(x) + C =
cot(x) + x + csc(x) + C

We feel pretty good about this since if we take a derivative we get:
1-csc²(x) - csc(x)cot(x) =
-cot(x) (1+ csc(x)) =
-cos(x) (1 + cos(x)) / sin²(x) =
-cos(x) (1 + cos(x)) / (1 - cos²(x)) =
-cos(x) / (1 - cos(x)) =
-1 / (sec(x) - 1) =
1 / (1 - sec(x))

Answer 3

try integration of cos x/cosx-1. See if it helps!
I have basically forgotten my math, thts the best i can do.