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This seems pretty simple to me but I can't get it.

$$\int \sin^2 x \cos^2 x dx$$

$$\int (1-\cos^2 x) \cos^2 x dx$$

I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an identity but this way seems easier, and I can't get an answer either way.

$$\int \cos^3 dx - \int \cos^ 5 x dx$$

I am not sure where to go from here, I don't know how to get the integral of $\cos^3 x$

  • Use the formula $2\cos^2(x)-1=\cos(2x)$. – guaraqe Jun 01 '12 at 15:01
  • With powers of sines and cosines, de Moivre's theorem or its equivalent formulations enable you to write a power of sine or cosine as a sum of sines and cosines (I understand this used to be used for multiplication, like logs). That gets the power into a form which is easy to integrate. It's worth trying out a few, so you can see the patterns which emerge. – Mark Bennet Jun 01 '12 at 15:41
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    After there are already several answers to your question, don't completely change it. Do a new question. This wastes the time of all the people who have already answered. Now, they either have an answer that doesn't match the problem, or they have to do a new solution (which would be somewhat similar but would take additional time). – GeoffDS Jun 01 '12 at 15:51
  • Does this answer your question? Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ – tryst with freedom May 13 '22 at 19:53

4 Answers4

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Write integrand as $(\sin x \cos x)^2 = (\frac{1}{2}\sin 2x)^2 $. Then use the following facts:

  • $\sin^2 2x = 1-\cos^2 2x$
  • $\cos^2 2x = \frac{1}{2}(\cos 4x +1)$

Note: The original question asked for the integral of $\sin^2 x \cos^2 x.$

Nana
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  • Dammit, I missed the trick –  Jun 01 '12 at 15:07
  • I don't see what is happening and how those are still equal expressions. –  Jun 01 '12 at 15:33
  • @Jordan: well, $\sin 2x = 2 \sin x \cos x$. so $\sin x \cos x =\ldots?$ Also, $\cos^2 x = \frac{1}{2}(\cos 2x + 1)$. These should help. If not let me know. – Nana Jun 01 '12 at 15:44
  • @Nana I do not see what is happening, or how $cos^ 2 x$ is equal to $1/2 cos2x+1$ –  Jun 01 '12 at 15:49
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An idea: $\,\,\displaystyle{\sin 2x=2\sin x\cos x\,\Longrightarrow \sin^2x\cos^2x=\frac{1}{4}\sin^22x}$.

Now just remember that $$\int\sin^2x\,dx=\frac{x-\sin x\cos x}{2}+C$$ so a little substitution solves the business.

DonAntonio
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  • Yes. Always simple hints work. – Mikasa Jun 02 '12 at 06:47
  • I do not like this because it requires an incredibly deep knowledge of trig or the memorization of very complex formulas that are very rarely used. –  Jun 02 '12 at 20:47
  • @Jordan "incredibly deep knowledge of trig" "very complex formulas". Again: try harder and don't give up. The formulas you are given are elementary. – Pedro Jun 02 '12 at 21:41
  • @PeterTamaroff You saying that they are elementary is just making me even more frustrated with this because I find them to be very complex, just trying to picture it all is very difficult, it probably takes a page of calculations to derive something like that. To me that is complex. –  Jun 02 '12 at 21:46
  • @Jordan I'll try and help you with those ones. Try not to put them on a pedestal, since it'll make things worse, seemingly unachievable, when they're not. – Pedro Jun 02 '12 at 21:48
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If you are dealing with powers of sine and cosine, you might find this reduction formula useful. Integrate by parts as follows. $u = \cos^{n-1}x$, $dv = \cos(x)\,dx$, $v = \sin(x)$, $du = (n-1)\cos^{n-1}(x)\sin(x)\,dx$ to to obtain

$$\int \cos^n(x)\,dx = \cos^{n-1}x\sin(x) -(n-1)\int \cos^{n-1}(x)\sin^2(x)\,dx$$

Apply the Pythagorean identity to get

$$ \int \cos^n(x)\,dx = \cos^{n-1}x\sin(x) -(n-1)\int \cos^{n-1}(x)(1 - \cos^2(x))\,dx $$

Break up the integral on the right and solve for $\int\cos^n(x)\,dx$ to see that $$n\int \cos^n(x)\,dx = \cos^{n-1}(x)\sin(x)-(n-1)\int \cos^{n-1}(x)\,dx$$ Finally, divide by $n$ and see that

$$\int \cos^n(x)\,dx = {1\over n}\cos^{n-1}(x)\sin(x)-{n-1\over n}\int \cos^{n-1}(x)\,dx $$ A similar arabesque is possible for the sine function. These reduction formulae may be applied repeatedly to tame powers of sine and cosine.

ncmathsadist
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  • You are using words and terms I am not familiar with. I do not know what arabesque is, and I find all the n-1 stuff pretty confusing. –  Jun 02 '12 at 20:37
  • @Jordan Try harder. Don't give up. – Pedro Jun 02 '12 at 21:39
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Write $\cos^{3}(x) = (1-\sin^{2}{x}) \cdot \cos{x}$ and put $t = \sin{x}$. And for your problem note that

  • $\cos{2x} = \cos^{2}(x) - \sin^{2}(x)= 2\cos^{2}(x)-1$. From here get the value of $\cos^{2}(x)$.

\begin{align*} \int \cos^{2}(x) \ dx &= \int \frac{1+\cos{2x}}{2} \ dx \\\ \int\cos^{4}(x) \ dx &= \int\biggl(\frac{1+\cos{2x}}{2}\biggr)^{2} \ dx \end{align*}