Find $\int \cos^4(x)dx$

Question

We have: $\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.$

Find $\int \cos^4x\ dx$ by using the formula twice

What I have so far is:

$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}\int \cos^{2} x\ dx$

Now we use the formula for $\int cos^{2} x\ dx$:

$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int \cos^{0} x\ dx$

$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int 1 \ dx$

$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}[ x ]$

$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{x}{2}$

Now plug this in to the $\int \cos^4 x\ dx$ equation above

$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}[\frac{1}{2} \cos x \sin x + \frac{x}{2}]$

$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{8}[\cos x \sin x + {x}]$

This is where I get stuck. I'm aware I could have used an identity for $\cos^2x$ but the question needs me to use the formula twice.

What precisely is the issue? You seem to be using the formula twice just fine. — Dennis Meng, Mar 04 '14 at 04:39
The answer I'm given is $\frac{3x}{8}+\frac{1}{4}sin(2x)+\frac{1}{32}sin(4x)+C$ — Mark, Mar 04 '14 at 04:43
@Mark So then, have you tried to check to see if the trig terms end up being the same from identities? You at least are matching up with the $3x/8$. — Dennis Meng, Mar 04 '14 at 04:45
Then there is a misunderstanding. That is obtained by using $\cos^2 t=\frac{\cos 2t+1}{2}$ twice. Another correct way of doing it, but it does not carry out the explicit instructions to use the Reduction Formula. — André Nicolas, Mar 04 '14 at 04:45
Another question about the same integral: http://math.stackexchange.com/questions/385938/integral-of-int-cos4xdx BTW you would get exactly the answer from your book using the approach mentioned in this answer. — Martin Sleziak, Mar 04 '14 at 09:16

score 1 · Accepted Answer · answered Mar 04 '14 at 05:53

1

You are right. Look at the alternate forms in WolframAlpha

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answered Mar 04 '14 at 05:53

Stefano

I guess that you could prove it by using the power reduction formulas http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula – Stefano Mar 04 '14 at 06:03
Thanks, I thought I could break it down further but, it seems like its an acceptable answer. – Mark Mar 05 '14 at 09:39

1 Answers1