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How do you find the integral of #int sin^n(x)# if m or n is an integer?

Answer 1

See below.

#d/(dx)(cos x sin^(n-1)x)=-sin^nx+(n-1)cos^2xsin^(n-2)x = # #=-sin^n x+(n-1)(1-sin^2x)sin^(n-2)x =# #= -sin^nx-(n-1)sin^n x+(n-1)sin^(n-2)x=# #=-nsin^n x+(n-1)sin^(n-2) x# so calling

#I_n = int sin^n x dx# we have

#I_n-((n-1)/n) I_(n-2)=-1/ncos x sin^(n-1)x# with

#I_0 = x# and #I_1 = -cos x#

So, with this recurrence equation we can build the response for a given #n#

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Answer 2

Scarlett Allison

To find the integral of (\int \sin^n(x) , dx) when (n) is an integer:

For odd (n): Use integration by parts.
For even (n): Use the half-angle identity for (\sin^2(x)) and then repeatedly apply this identity until the expression becomes in terms of (\sin(x)) to a power that is odd, and then proceed with integration by parts.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.