Integration of an algebraic and trigonometrical function

Arkavo
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Homework Statement


x
∫((x^n)sinx)dx=0.75(\pi^2-8) find n if n is a single digit positive integer
0

Homework Equations



∫uv.dx=u∫v.dx-∫(u'.∫v.dx).dx

The Attempt at a Solution



i tried putting n=1 or 2 but didnt get the result
i somehow don't think that's the method
help please
 
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Arkavo said:

Homework Statement


x
∫((x^n)sinx)dx=0.75(\pi^2-8) find n if n is a single digit positive integer
0

Homework Equations



∫uv.dx=u∫v.dx-∫(u'.∫v.dx).dx

The Attempt at a Solution



i tried putting n=1 or 2 but didnt get the result
i somehow don't think that's the method
help please

Hi Arkavo!

Are you sure about the integration limits? Can you please recheck the question?
 
ok no the x is replaced by pi/2
sorry
 
Arkavo said:

Homework Statement


x
∫((x^n)sinx)dx=0.75(\pi^2-8) find n if n is a single digit positive integer
0

Homework Equations



∫uv.dx=u∫v.dx-∫(u'.∫v.dx).dx

The Attempt at a Solution



i tried putting n=1 or 2 but didnt get the result
i somehow don't think that's the method
help please

You need to show your work (as required by PF rules); but first, you need to clarify your question. Do you mean that you want to solve the equation
\int_0^x t^n \sin(t) \, dt = \frac{3}{4} \left( \pi^2 - 8 \right)
to find x? Different small integer values of n give different solutions x, so you need to spell out in more detail what you want.

What was the value of your integral for n = 1 and for n = 2?
 
n=3

NR
 
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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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