Is this process incorrect (integral)
- Thread starter cwerry
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topsquark
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[imath]\int x^n ~ dx = \dfrac{1}{n + 1} x^{n + 1}[/imath] but the integrand of [imath]\int sin^7(x) ~ dx[/imath] is not of the form [imath]x^n[/imath]. On a quick check your work up to the line where you got the integral in terms of just powers of sines is good. To do each of the sine integrals you need to either use a reduction formula (ie. a table) or you can use [imath]sin^2(x) = 1 - cos^2(x)[/imath] to cut down on some of the powers of sine, then use a substitution u = cos(x).
-Dan
-Dan
[imath]\sin^7{x} \cos^4{x} = \sin {x} \sin^6 {x} \cos^4 {x}\\
\hspace12ex= \sin{x} (1-\cos^2{x})^3 \cos^4{x}\\
\hspace12ex=\sin{x} \cos^4{x} -3\sin{x} \cos^6{x} +3 \sin{x} \cos^8{x} -\sin{x}\cos^{10}{x}[/imath]
Each of these can be easily integrated.
[imath]\int \sin{x}\cos^n{x} \text{ dx}=-\dfrac{\cos^{n+1}{x}} {n+1} + c[/imath]
Each of these can be easily integrated.
[imath]\int \sin{x}\cos^n{x} \text{ dx}=-\dfrac{\cos^{n+1}{x}} {n+1} + c[/imath]
When you do an integration, you should always check it by differentiating.
You have:
[imath]\int \sin^7{x} \text{ dx} = -\frac{1}{8} \cos^8{x} + c\\ \text{ }\\ \text{but }\frac{d}{dx} \left( -\frac{1}{8}\cos^8{x}\right)= -\frac{1}{8} \left( 8\cos^7{x} \sin{x} \right) \text{ not } \sin^7{x}[/imath]
So,in fact, contrary to appearances [imath]\int \sin{x} \cos^7{x} \text{ dx}[/imath] is easy,
[imath]\int\sin^7{x} \text{ dx}[/imath] is harder.
That's why, when I saw your question, noticing that there was an odd number of sines, I kept one and converted the others to cosines, knowing that I would then be able to integrate easily.
You have:
[imath]\int \sin^7{x} \text{ dx} = -\frac{1}{8} \cos^8{x} + c\\ \text{ }\\ \text{but }\frac{d}{dx} \left( -\frac{1}{8}\cos^8{x}\right)= -\frac{1}{8} \left( 8\cos^7{x} \sin{x} \right) \text{ not } \sin^7{x}[/imath]
So,in fact, contrary to appearances [imath]\int \sin{x} \cos^7{x} \text{ dx}[/imath] is easy,
[imath]\int\sin^7{x} \text{ dx}[/imath] is harder.
That's why, when I saw your question, noticing that there was an odd number of sines, I kept one and converted the others to cosines, knowing that I would then be able to integrate easily.