Finding the Indefinite Integral: Can Multiplying a Constant Change the Solution?

Scholar1
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I have posted my attempt and the problem above. Please help!

Thanks in advance!
 
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Scholar1 said:
IMG_8357.jpg


I have posted my attempt and the problem above. Please help!

Thanks in advance!
It's hard to mark up an image. :frown:

The factor of (1/2) multiplies not only ∫ 1 dx but also ∫ -cos 6x dx, which you omitted from the second integral. :frown:

After you have found the correct integrals, you can use the double angle formulas to convert these from trig functions in 6x to trig functions in 3x. :wink:
 
So if I put - (1/12) in front of the sin6x it would be correct?
 
Scholar1 said:
So if I put - (1/12) in front of the sin6x it would be correct?
Yesss...multiplication distributes over addition and subtraction.
 
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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