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Solving the integral of $\cos^2x\sin^2 x$:

My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that

$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x))$$ Rewriting all the steps:

$$\int \sin ^2\left(x\right)\cos ^2\left(x\right)dx=\frac{1}{8}\left(x-\frac{1}{4}\sin \left(4x\right)\right)+k, \,\, k\in \Bbb R \tag 1$$

Is there another method to solve this integral (1)?

Sebastiano
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    Of course there are other ways to do it, but isn't this way especially neat and convenient? – David K Jun 04 '22 at 15:41
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    in general, for an integral of the form $$\int \cos^n x\sin^m x,dx$$ order reduction using double angle formula etc. is the best approach for $n=m$ – Henry Lee Jun 04 '22 at 15:42
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    @DavidK If you already have the results for $\int \sin^2 x dx$ and $\int \sin^4 x dx$ then you can do a neater job by using $\cos^2 x = 1 - \sin^2 x$. – Prime Mover Jun 04 '22 at 15:57
  • @DavidK so wots yore point? – Prime Mover Jun 04 '22 at 18:24
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    @PrimeMover Well, I guess I could get all sore about (apparently) being contradicted on what I thought was mostly a compliment to the OP (who I still think did a very nice work on the integral even before asking the question), but since last visiting this page I went out in a kayak and saw geese sleeping on one leg and muskrats swimming and everything seems good. So yes, it's good to know about power reduction (for more reasons than just this) and of course it is good to see an integral solved different ways. So, useful question and useful answer. – David K Jun 04 '22 at 21:57

2 Answers2

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You can excpress $\cos^2 x$ as $1 - \sin^2 x$ and hence your primitive becomes: $$\int \sin^2 x d x - \int \sin^4 x dx$$

which comes to the same answer in the end.

Full working:

$$\int \sin^2 a x \cos^2 a x \, \mathrm d x$$

$$=\int \sin^2 a x (1 - \sin^2 a x) \, \mathrm d x$$

$$=\int \sin^2 a x \, \mathrm d x - \int \sin^4 a x \, \mathrm d x$$

$$=\frac x 2 - \frac {\sin 2 a x} {4 a} - \int \sin^4 a x \, \mathrm d x + C$$

$$= \frac x 2 - \frac {\sin 2 a x} {4 a} - \left ({\frac {3 x} 8 - \frac {\sin 2 a x} {4 a} + \frac {\sin 4 a x} {32 a} }\right) + C$$

$$= \frac x 8 - \frac {\sin 4 a x} {32 a} + C$$

In the above:

$$\int \sin^4 a x \, \mathrm d x = \int \left( {\frac {3 - 4 \cos 2 a x + \cos 4 a x} 8}\right) \, \mathrm d x$$

using the usual power reduction formula.

Prime Mover
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  • Please can you extend your answer? Thank you very much. – Sebastiano Jun 04 '22 at 15:43
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    @Sebastiano Sorry, I thought you were just interested in another method of solution. I assumed that, given that you have been given a method, that you can then work on the solution from there. Apologies, will expand it in due course. – Prime Mover Jun 04 '22 at 15:49
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Integrate by parts \begin{align} \int \cos^2x\sin^2 x \ dx =& -\frac1{16}\int\tan 2x\ d(\cos^2 2x)= -\frac1{16}\tan2x\cos^22x+\frac18 x+C \end{align}

Quanto
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  • This still needs some work to be done to reduce that final expression into the same form as that given by OP. The integration itself is far from obvious and could do with being worked through. – Prime Mover Jun 05 '22 at 06:54