Antiderivative of 1.4x*cos(x^1.9)

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Question

f'(x) = 1.4x*cos(x^1.9)
Find f(x)

Attempt
Ok, first of all, I'm really bad at Calculus, so bear with me >__<
I figured to find the antiderivative of the thing is equivalent to
\int\left1.4x*cos\leftx^{1.9}\right\right)dx

I've tried simple substitution (u = x^1.9) and that obviously doesn't work, and I tried integration by parts, but I don't think it's working
If I had dv = cos(x^1.9)dx
I'm unsure if I can find v when x is x^1.9

Help will be appreciated! Thanks!
 
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momogiri said:
Question

f'(x) = 1.4x*cos(x^1.9)
Find f(x)

Attempt
Ok, first of all, I'm really bad at Calculus, so bear with me >__<
I figured to find the antiderivative of the thing is equivalent to
\int\left1.4x*cos\leftx^{1.9}\right\right)dx

I've tried simple substitution (u = x^1.9) and that obviously doesn't work, and I tried integration by parts, but I don't think it's working
If I had dv = cos(x^1.9)dx
I'm unsure if I can find v when x is x^1.9

Help will be appreciated! Thanks!

Your doing the math lab arn't you? I have the same problem except my f'(x) = 1.7x*sin(x^1.8)
 
You have two different Integrals ... which is it?

\int1.4x\cos{x^{1.9}}dx

or

\int1.4x\cos^{1.9}xdx

Does the power belong to cosine or the variable x?
 
Have you tried Taylor expansion? That's all I can think of now.
 
If you're doing the mathlab for a UBC math course *cough*, then use the spreadsheet.
 
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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