Finding integrand of trig function.

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SUMMARY

The integral of the function \(\int \sin^4(\pi\theta) \cos^3(\pi\theta) d\theta\) was analyzed, revealing a common approach using substitution. The substitution \(u = \sin(\pi x)\) and \(du = \frac{1}{\pi} \cos(\pi x) dx\) was employed, leading to an integral transformation. The final expression was derived as \(\frac{u^5}{5\pi} + \frac{u^7}{7\pi} + c\), with noted errors in multiplication signs and variable substitution. Correcting these errors is essential for accurate results.

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Homework Statement


[tex]\int sin^4({\Pi\theta}) cos^3({\Pi\theta}) d\theta[/tex]


The Attempt at a Solution


let x = theta to make things simpler in text
let u = sin (pi x)
du = (1/pi) cos (pi x) dx

integral sin^4(pi x) cos^3(pi x) dx
= integral sin^4(pi x) cos^2(pi x) cos (pi x) dx
= integral sin^4(pi x) [1-sin^2 (pi x)] cos (pi x) dx
= integral sin^4(pi x) cos(pi x) . sin^6(pi x) cos(pi x) dx
= (1/pi) integral u^4 + u^6 du
= (1/pi) (u^5/5+u^7/7) + c
= u^5/5pi + u^7/7pi + c.

I know the answer is wrong, but I think the method I used was following the textbook. Can someone point out where I was wrong?

Thank you! :)
 
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intermu said:

Homework Statement


[tex]\int sin^4({\Pi\theta}) cos^3({\Pi\theta}) d\theta[/tex]

The Attempt at a Solution


let x = theta to make things simpler in text
let u = sin (pi x)
du = (1/pi) cos (pi x) dx

integral sin^4(pi x) cos^3(pi x) dx
= integral sin^4(pi x) cos^2(pi x) cos (pi x) dx
= integral sin^4(pi x) [1-sin^2 (pi x)] cos (pi x) dx
= integral sin^4(pi x) cos(pi x) . sin^6(pi x) cos(pi x) dx

The Multiplication Sign there is incorrect. :(

= (1/pi) integral u^4 + u^6 du
= (1/pi) (u^5/5+u^7/7) + c
= u^5/5pi + u^7/7pi + c.

Wrong sign. :(

Apart from those little (and "silly") mistakes, everything looks just fine to me. :)

Ah, I forgot to mention that you also need to switch u back to x. Since the problem is in terms of x, so it's more natural to have the result displayed in terms of x, too. :)
 
Last edited:

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