Calculus Problems part I

1. Jan 11, 2006

Nimmy

Can someone show me the steps on finding:

Note i dont have the answer for the last two questions. Im so confused help :surprised

1. ∫1/cscx-1

2
_________ - x
cot (x/2)-1

2. ∫1/(1-sin x)

2
_________
cot (x/2)-1

3. ∫1/(1-e^-x)

4. ∫cos x ln x

2. Jan 12, 2006

VietDao29

You should note that:
∫1/cscx-1 is not the same as ∫dx/(cscx-1).
When seeing these kinds of integrals (i.e: trig integral), generally, what you should do is to use u-substitution (u = tan(x / 2)).
Let $$u = \tan \left( \frac{x}{2} \right) \Rightarrow du = \frac{dx}{2 \cos ^ 2 \left( \frac{x}{2} \right)} = \frac{dx (1 + u ^ 2)}{2} \Rightarrow dx = \frac{2du}{1 + u ^ 2}$$
You'll also have:
$$\sin x = \frac{2u}{1 + u ^ 2}$$
$$\int \frac{dx}{\csc x - 1} = \int \frac{\sin x \ dx}{1 - \sin x} = \int -\frac{-\sin x \ dx}{1 - \sin x} = \int -\frac{1 - \sin x - 1 \ dx}{1 - \sin x}$$
$$= -\int dx + \int \frac{dx}{1 - \sin x} = -\int dx + \int \frac{2du}{(1 + u ^ 2) \left( 1 - \frac{2u}{1 + u ^ 2} \right)} = -\int dx + \int \frac{2du}{(1 + u ^ 2) \left( 1 - \frac{2u}{1 + u ^ 2} \right)}$$.
Can you go from here?
It can be done exactly the same as the first one.
To do some kinds of integrals like this, you can use the substitution u = ex.
This can be done by integrating by parts. Looking closely at that, what should be u, and what should dv?
You should note that:
$$\int \frac{\sin x}{x} dx = \mbox{si} (x) + C$$.
Can you go from here?
-------------------
If you get stuck somewhere, just should it out.

Last edited: Jan 12, 2006
3. Jan 12, 2006

HallsofIvy

Staff Emeritus
Strictly speaking $\int1/csc x- 1$ means $\int\frac{1}{csc x}-1 dx$ which is the same as $\int sin(x)-1 dx$ which is easy. I think you meant $\int\frac{1}{csc x- 1}dx$.
That is the same as $\int\frac{sin x}{1- sin x}dx$