Solving Integrals with Trigonometric Functions

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Homework Statement


∫ x sinx cosx dx =


Homework Equations


note that sinx cosx = (1/2) sin 2x


The Attempt at a Solution


1/2) ∫ x sin 2x dx =
(1/2) [(-1/2)x cos 2x - ∫ (-1/2)cos 2x dx] =
(1/2) [(-1/2)x cos 2x +(1/2) ∫ cos 2x dx] =
(-1/4)x cos 2x +(1/4) ∫ cos 2x dx =
(-1/4)x cos 2x +(1/4)(1/2) sin 2x + c =
(-1/4)x cos 2x +(1/8) sin 2x + c =
:bugeye:
 
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chapsticks said:

Homework Statement


∫ x sinx cosx dx =


Homework Equations


note that sinx cosx = (1/2) sin 2x


The Attempt at a Solution


1/2) ∫ x sin 2x dx =
(1/2) [(-1/2)x cos 2x - ∫ (-1/2)cos 2x dx] =
(1/2) [(-1/2)x cos 2x +(1/2) ∫ cos 2x dx] =
(-1/4)x cos 2x +(1/4) ∫ cos 2x dx =
(-1/4)x cos 2x +(1/4)(1/2) sin 2x + c =
(-1/4)x cos 2x +(1/8) sin 2x + c =
:bugeye:

That looks just fine. What's the question?
 
It seems like you have an answer. So what's the issue?
 
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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