Solve Integral Calculus: Find a Value for a

planauts
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Hi, I have another question that I am having trouble with:

Homework Statement


Find the value of a such that:
\int_{0 }^{a } cos^2 (x) dx

The Attempt at a Solution

\int_{0}^{a}\frac{1}{2} (1+cos (2x))

\frac{x}{2} + \frac{(sin(2x)}{4}

\frac{a}{2} + \frac{sin(2a)}{4} - 0 - 0 = 0.740

\frac{a}{2} + \frac{sin(2a)}{4} = 0.740

2a + sin(2a) = 2.96

I am stuck after this...How would I solve the trig equation?

Maybe...m + sin(m) = 2.96
where m = 2a
But I don't think that will get me anywhere :-/

Thanks for your help!
 
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This equation can not b solved elementary.
Did you post the whole question?
Also, your question isn't complete. Such the integral equal what?
 
AfterSunShine said:
This equation can not b solved elementary.
Did you post the whole question?
Also, your question isn't complete. Such the integral equal what?

This is what the question says:

Find the value of a such that \int_{0 }^{a } cos^2 (x) dx<br />. Give your answer to 3 decimal places.

The answer is: a = 1.047

EDIT:
The answer does work but I don't know how my teacher got it. :-/
<br /> 2(1.047) + sin(2(1.047)) = 2.96<br />
 
I would use Newton's method to find the numerical value of the root of <br /> 2a + sin(2a) = 2.96<br />.
 
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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