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Integral of cos^5x

\int \cos^5(x) \mathrm{d}x

I know there's a standard way of approaching even and odd trig questions but I can't remember it. I think you start by factorising so it's reduced to an even trig expression:

cos^5x = cosx(cos^4x)

(cos2x + 1)/2 = cos^2x

(cos^22x + 2cos2x + 1)/4 = cos^4x

But then substituting this back isn't really leading anywhere.

Reply 1

DFranklin

18

Funnily, even powers are considerably harder than odd ones.

For an odd power as here, write

cos^5 x = cos x cos^4 x

= cos x(1-sin^2 x)^2

= cos x(1 - 2 sin^2 x + sin^4 x).

Now substitute t = sin x (by sight/recognition if you wish).

For an even power, I'd use a reduction formula.

Reply 2

Kolya

17

Writing

\cos ^4 x = (1 - \sin ^2x)^2

works.

Reply 3

Swayum

OP

18

Ah right, thanks.

What's a reduction formula?

Reply 4

Kolya

17

Swayam

Ah right, thanks.

What's a reduction formula?

Where an integral is reduced to one of a lower power that is easier to compute. For example, it is possible to find a reduction formula for

\displaystyle \int \cos ^n x \ dx

(You could give it a go, if you want, and I will check it for you.)

Reply 5

applicant1002

Another way to do things (if you're familiar with it) would be to write cos^5x in terms of complex exponentials, multiply out and then recombining to get an integrand of the sum 1/16(cos5x + 5cos3x + 10cosx).

a bit easier in my opinion once you've done a bit of FP2

Reply 6

DFranklin

18

Original post by applicant1002

Another way to do things (if you're familiar with it) would be to write cos^5x in terms of complex exponentials, multiply out and then recombining to get an integrand of the sum 1/16(cos5x + 5cos3x + 10cosx).

a bit easier in my opinion once you've done a bit of FP2

Compare:

\cos^5x = \left(\dfrac{e^{ix} + e^{-ix}}{2}\right)^5 = \dfrac{1}{32}(e^{5ix} + \binom{5}{4}e^{i3x} + \binom{5}{3} e^{ix} + \binom{5}{3} e^{-ix} + \binom{5}{1} e^{-i3x} + e^{-5ix})

=\dfrac{1}{32} (2 \cos 5x + 10 \cos 3x + 20 \cos x) = \dfrac{1}{16} (\cos 5x + 5 \cos 3x + 10 \cos x)

Integrating, we get

\dfrac{1}{16} \left( \dfrac{\sin 5x}{5} + \dfrac{5 \sin 3x}{3} + 10 \sin x \right)

= \dfrac{1}{240} (3 \sin 5x + 5 \sin 3x + 150 \sin x)

(assuming I haven't made a mistake somewhere).

Doing it my way:

\int \cos^5 x \, dx = \int \cos x (1-\sin^2 x)^2\, dx = \int \cos x (1 - 2 \sin^2 x + \sin^4 x)\,dx

=\sin x - \dfrac{2 \sin^3 x}{3} + \dfrac{\sin^5 x}{5} = \dfrac{1}{15} (15 \sin x - 10 \sin^3 x + 3 \sin^5 x)

Arguably your method is easier than a reduction formula; it depends a bit on the limits, really.

Reply 7

raheem94

13

Original post by DFranklin

.............

Isn't the method given in the below image easier:

Reply 8

DFranklin

18

Seeing as I got there in about a quarter of the space, I say no.

Reply 9

applicant1002

That's the method I was using with the gaps filled in. Once you've done it a number of times you can do it without a lot of those steps. I did the integral in 3 steps on one line, so it would be quicker in my opinion, and offers a more general approach to these sorts of integrals. Also it gives you a function that is easy to differentiate so you can check you're answer as well.

It's a matter of preference really.

Reply 10

Zahurul Islam

1

but when the limit is o to π then??

Reply 11

ghostwalker

17

Original post by Zahurul Islam

but when the limit is o to π then??

It's general policy not to resurrect old threads - start a new one; see forum guidelines, sticky thread at top of fourm -, but to answer your question.

If you can't see it straight off, then one method, is substitution, let

x=u+\pi/2

This converts the integral to

\displaystyle\int_{-\pi/2}^{\pi/2}-\sin^5 u \;du

.

This is an odd function integrated over an interval symmetrical about 0, and hence integrates to 0.

Note: The original question was for an indefinite integral.

(edited 6 years ago)

Reply 12

username2389033

4

Integral of cos^5x

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