How can I integrate trig functions using u-substitution and n-substitution?

marc017
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This isn't homework, It is just book problems that I am practicing, I am checking some answers with wolfram and others with the book answers.

Homework Statement



<br /> \begin{align}<br /> \int \frac{sin(w)\,dw}{\sqrt{1-cos(w)}}\\<br /> \end{align}<br />

Homework Equations



I used u substitution... Not sure if I approached this problem the correct way

The Attempt at a Solution



<br /> \begin{align}<br /> \int \frac{sin(w)\,dw}{\sqrt{1-cos(w)}}\\<br /> \end{align}<br />

Using U sub... U = cos(w), du = -sin(w)

<br /> \begin{align}<br /> - \int \frac{\,du}{\sqrt{1-u}}\\<br /> \end{align}<br />

Using n sub... n=1-u, dn = -1

<br /> \begin{align}<br /> \int \frac{\,du}{\sqrt{n}} = 2\sqrt{(1-cos(w))} + C\\<br /> \end{align}<br />
 
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marc017 said:
This isn't homework, It is just book problems that I am practicing, I am checking some answers with wolfram and others with the book answers.

Homework Statement



<br /> \begin{align}<br /> \int \frac{sin(w)\,dw}{\sqrt{1-cos(w)}}\\<br /> \end{align}<br />


Homework Equations



I used u substitution... Not sure if I approached this problem the correct way

The Attempt at a Solution



<br /> \begin{align}<br /> \int \frac{sin(w)\,dw}{\sqrt{1-cos(w)}}\\<br /> \end{align}<br />

Using U sub... U = cos(w), du = -sin(w)

<br /> \begin{align}<br /> - \int \frac{\,du}{\sqrt{1-u}}\\<br /> \end{align}<br />

Using n sub... n=1-u, dn = -1

<br /> \begin{align}<br /> \int \frac{\,du}{\sqrt{n}} = 2\sqrt{(1-cos(w))} + C\\<br /> \end{align}<br />

Looks good. One nice thing about these types of problems is that you can check them yourself. If your answer is correct, you should be able to differentiate it and get the integrand.

As for your substitution, what you did is OK, but you can kill two birds with one stone by letting u = 1 - cos(w). Then du = sin(w)dw.
 
Mark44 said:
Looks good. One nice thing about these types of problems is that you can check them yourself. If your answer is correct, you should be able to differentiate it and get the integrand.

As for your substitution, what you did is OK, but you can kill two birds with one stone by letting u = 1 - cos(w). Then du = sin(w)dw.

Thank you.. I was checking the integrals on wolfram but it seems to work much better if I take my answer and try to get the integral. And I can't believe I didn't think of the 1-cos(w) substitution :redface:

I just found this forum and you guys have been a lot of help! Maybe one day i will be good enough to answer other people's questions.
 
marc017 said:
Thank you.. I was checking the integrals on wolfram but it seems to work much better if I take my answer and try to get the integral.
...
Speaking of using WolframAlpha, their result for this integration is ##\displaystyle \ \frac{4 \sin^2(x/2)}{(1-\cos(x))^{3/2}}+C \ .##

By the way, this is equivalent to your answer.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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