How Do I Solve These Calculus Problems?

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In summary: So the solution to the first integral is \int\frac{sin x}{1-sin x}dx= -\int\frac{1-sin x}{1-sin x}dx= -\int dx=-x+C. In summary, the conversation discusses finding the steps for solving four different integrals, with the last two being more difficult and needing special techniques. The first two integrals involve using u-substitution, while the third involves using the substitution u = e^x. The fourth integral requires integrating by parts. The last two integrals are not solved in detail, but hints are given for how to approach them.
  • #1
Nimmy
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Can someone show me the steps on finding:

Note i don't have the answer for the last two questions. I am so confused help

1. ∫1/cscx-1

It says the answer is:

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

2. ∫1/(1-sin x)
It says the answer is:

2
_________
cot (x/2)-1

3. ∫1/(1-e^-x)

4. ∫cos x ln x
 
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  • #2
Nimmy said:
Can someone show me the steps on finding:
Yes, we can help you, but we won't show you the step-by-step solution! https://www.physicsforums.com/showthread.php?t=28
Nimmy said:
Note i don't have the answer for the last two questions. I am so confused help

1. ∫1/cscx-1

It says the answer is:

2
_________ - x
cot (x/2)-1
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 [tex]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}[/tex]
You'll also have:
[tex]\sin x = \frac{2u}{1 + u ^ 2}[/tex]
[tex]\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}[/tex]
[tex]= -\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)}[/tex].
Can you go from here?
Nimmy said:
2. ∫1/(1-sin x)
It says the answer is:

2
_________
cot (x/2)-1
It can be done exactly the same as the first one.
Nimmy said:
3. ∫1/(1-e^-x)
To do some kinds of integrals like this, you can use the substitution u = ex.
Nimmy said:
4. ∫cos x ln x
This can be done by integrating by parts. Looking closely at that, what should be u, and what should dv?
You should note that:
[tex]\int \frac{\sin x}{x} dx = \mbox{si} (x) + C[/tex].
Can you go from here?
-------------------
If you get stuck somewhere, just should it out.
 
Last edited:
  • #3
Strictly speaking [itex]\int1/csc x- 1[/itex] means [itex]\int\frac{1}{csc x}-1 dx[/itex] which is the same as [itex]\int sin(x)-1 dx[/itex] which is easy. I think you meant [itex]\int\frac{1}{csc x- 1}dx[/itex].
That is the same as [itex]\int\frac{sin x}{1- sin x}dx[/itex]
 

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The two main branches of Calculus are differential calculus and integral calculus. Differential calculus deals with the rates of change of a function, while integral calculus deals with the accumulation of these changes over a given interval.

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