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

Answer 1

See the explanation for one way to do these.

Even #n# For #n=2k# use #cos(2x) = 2cos^2x-1# to get the power reduction identity:
#cos^2x = 1/2(1+cos(2x))#

So

#(cos^2x)^k = (1/2(1+cos(2x)))^k#
Expand the power #k# and reduce any
Odd #n# For #n=2k+1# rewrite as
#cos^(2k+1)x = (cos^2x)^k cosx#
# = (1-sin^2x)^k cosx#
Now use # cos(2x) = 1-2sin^2x# to get the power reduction identity:
#sin^2x = 1/2(1-sin(2x))#.
Us power reduction, expanding the power #k#, then power reduction again and repeat as needed to get an integral of the form #int(k+a_nsin^(b_n) u + a_(n-1)sin^(b_2) u + * * * +a_1sin^(b_1)u) cosu du#

Next, use substitution to integrate each term one by one.

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Answer 2

To find the integral of (\int \cos^n(x) , dx) where (n) is a positive integer, you can use the method of reduction formulas. The formula for (n > 1) is:

[ \int \cos^n(x) , dx = \frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} \int \cos^{n-2}(x) , dx ]

And for (n = 1):

[ \int \cos(x) , dx = \sin(x) + C ]

Where (C) is the constant of integration. You can repeatedly apply the reduction formula until the integral becomes manageable.

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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.

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.

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.

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.

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