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Help with a question
- Thread starter goldgold
- Start date
D
Deleted member 4993
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So some how I was supposed to get from here
-log(cot(x)+csc(x))+constant
To here
= log(sin(x/2))-log(cos(x/2))+constant
The problem was
∫dx/sinx
I tried allot of trigonometric function, but I can't make the pass between them
It is fairly simple trigonometric manipulation:
cot(x) + csc(x) = cos(x)/sin(x) + 1/sin(x) = [cos(x)+1]/sin(x)
Now use the fact that cos(2Θ) = 2cos2(Θ) -1
and continue....
ֺ
Hi gold^2
You already rewrote the cotangent and cosecant functions each in terms of sine and cosine functions, followed by combining those ratios into a single ratio, yes?
The negative sign in front of the first log means multiplication by -1; use the property of logarithms to move that factor of -1 into the exponent position.
After doing that, use the property of negative exponents to change the ratio to its reciprocal.
Next, you need to change the inputs to sine and cosine (from x to x/2). Use the double-angle formulas for that.
sin(2u) = 2 sin(u) cos(u)
cos(2u) = 2 cos(u)^2 - 1
Let u = x/2; so that 2u = x
After some simplification and cancelling of common factors, we have log[sin(x/2)/cos(x/2)^2] + C
Finish, by applying the property which changes a logarithm of a ratio into a difference of logarithms.
If you cannot follow all of these steps, then please show how far you got, and we'll continue from there.
Cheers :cool:
ֺ
ֺsupposed to get from here
-log(cot(x) + csc(x)) + constant
To here
log(sin(x/2)) - log(cos(x/2)) + constant
Hi gold^2
You already rewrote the cotangent and cosecant functions each in terms of sine and cosine functions, followed by combining those ratios into a single ratio, yes?
The negative sign in front of the first log means multiplication by -1; use the property of logarithms to move that factor of -1 into the exponent position.
After doing that, use the property of negative exponents to change the ratio to its reciprocal.
Next, you need to change the inputs to sine and cosine (from x to x/2). Use the double-angle formulas for that.
sin(2u) = 2 sin(u) cos(u)
cos(2u) = 2 cos(u)^2 - 1
Let u = x/2; so that 2u = x
After some simplification and cancelling of common factors, we have log[sin(x/2)/cos(x/2)^2] + C
Finish, by applying the property which changes a logarithm of a ratio into a difference of logarithms.
If you cannot follow all of these steps, then please show how far you got, and we'll continue from there.
Cheers :cool:
ֺ
Last edited:
So some how I was supposed to get from here
-log(cot(x)+csc(x))+constant
to here
= log(sin(x/2))-log(cos(x/2))+constant
The problem was
∫dx/sinx
goldgold, you and certain other users are missing required absolute value bars.
And it is log to the base e, so I am explicitly showing the base, instead of the sometimes ambiguous word "log."
\(\displaystyle -log_e\bigg|csc(x) \ + \ cot(x)\bigg| \ + \ C \ =\)
\(\displaystyle log_e\bigg|sin\bigg(\dfrac{x}{2}\bigg)\bigg| \ - \ log_e\bigg|cos\bigg(\dfrac{x}{2}\bigg)\bigg| \ + \ C \)