What technique would be suitable to solve this: $\int \sin ^{5}\left( x^{2}\right) \left( x\cos \left(x^{2}\right)\right)\mathrm{d}x$

I think integration by parts might work but I'm now sure. Thanks very much.

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what is 'sen'? Is it a typo? – Willie Wong Jul 5 '11 at 16:44
I assume it is 'sin'. At least so is in Portuguese. – Américo Tavares Jul 5 '11 at 16:52
And so is in Spanish. – Adrián Barquero Jul 5 '11 at 17:04

I assume the integral is

$$\int \sin ^{5}\left( x^{2}\right) \left( x\cos \left( x^{2}\right) \right) \mathrm{d}x.$$

Edited 2: Let $f(x)=\sin ^{6}\left( x^{2}\right)$. Then

$$f^{\prime }\left( x\right) =6\ \sin ^{5}(x^{2})\left( \cos x\right) \cdot 2x=12\ \sin ^{5}\left( x^{2}\right) \left( x\cos \left(x^{2}\right)\right) .$$

So, since $\int f'(x)\mathrm{d}x=f(x)+C$, we get

$$\begin{eqnarray*}\int \sin ^{5}\left( x^{2}\right) \left( x\cos \left( x^{2}\right) \right) \mathrm{d}x&=&\dfrac{1}{12}\int 12\ \sin ^{5}\left( x^{2}\right) \left( x\cos \left(x^{2}\right)\right)\mathrm{d}x \\ &=&\frac{1}{12}\sin ^{6}(x^{2})+C.\end{eqnarray*}$$

Comment: I thought of the function $f(x)=\sin ^{6}\left( x^{2}\right)$ because I saw the fifth power factor $\sin ^{5}\left( x^{2}\right)$ in the integrand, and I knew that by differentiation I would get this factor times the derivative of the base ($\sin \left( x^{2}\right)$), which should contribute to a $\cos$ factor. It happened that by luck I got apart from the $12$ factor the integrand. I mean that if the integrand were e.g. $\sin ^{5}\left( x^{2}\right) \left( \cos \left( x^{2}\right) \right)$ I would not succeed.

Added: Alternatively we can use the substitution $u=\sin (x^{2})$ recommended by The Chaz. We then have $\mathrm{d}x=\dfrac{1}{2x\left( \cos \left( x^{2}\right)\right) }\mathrm{d}u$ and

$$\begin{eqnarray*} \int \sin ^{5}\left( x^{2}\right) \left( x\cos \left( x^{2}\right) \right) \mathrm{d}x &=&\int u^{5}\left( x\cos \left( x^{2}\right) \right) \frac{1}{2x\left( \cos \left( x^{2}\right) \right) }\mathrm{d}u \\ &=&\frac{1}{2}\int u^{5}\mathrm{d}u=\frac{1}{2}\cdot \frac{u^{6}}{6}+C=\frac{u^{6}}{12}+C \\ &=&\frac{\sin ^{6}(x^{2})}{12}+C. \end{eqnarray*}$$

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Or let $u= sin(x^2)$... – The Chaz 2.0 Jul 5 '11 at 17:07
@The Chatz: You are right! – Américo Tavares Jul 5 '11 at 18:56
@The Chatz: Thanks! I added your substitution to my answer. – Américo Tavares Jul 5 '11 at 19:20
Glad you like it! Does "chatz" mean something in Portuguese? "Chaz" is a nickname for Charles, my legal name. – The Chaz 2.0 Jul 5 '11 at 19:36
@The Chaz: No. The diminuitives Charley and Charlie translate to "Carlitos" or "Carlinhos". And Charles is Carlos. I don't know why I wrote "The Chatz", perhaps because I thought of the German word "Schatz". – Américo Tavares Jul 5 '11 at 20:11

I will do the problem in not the most efficient way. We want $$\int\sin^5(x^2)(x\cos(x^2))\,dx.$$ In our expression almost all of the parts are functions of $x^2$. It would be nicer not to have those $x^2$ "inside," they complicate things.

A more or less immediate reaction is to notice that the derivative of $x^2$ is (almost) sitting in our expression. So it seems sensible to let $u=x^2$. Then $du=(2x)\,dx$, or equivalently, $xdx=(1/2)\,du$.

Now make the substitution, mechanically.

We obtain $$\int (1/2) (\sin^5 u)(\cos u)\,du.$$ It is beginning to look better, and often what looks better is better.

Now we notice that the derivative of $\sin u$, namely $\cos u$, is sitting in our expression. This makes it tempting to make the substitution $v=\sin u$. Then $dv=\cos u \,du$, and making the mechanical substitution we obtain $$\int (1/2)v^5\, dv.$$ The integral is now immediate. We get $$\frac{1}{12}v^6 +C.$$ Now substitute back, to get an expression in $x$.

With experience, as we develop a better "eye" for things, the two substitutions that I made can be collapsed into one, and maybe one can even write down the integral directly.

Comment: The integration by parts idea that you had is motivated by integrals like $\int x\sin x\, dx$, where integration by parts works nicely. My impulse is, if possible, to try to get rid of the "$x^2$" that are deep inside our expression. After that initial cleaning up has been done, we might need to do an integration by parts. But initial tidying, if it can be done, is often a good idea. In general, the more tidy things look, the more likely we are to succeed.

Added Comment: There is a fair chance that you were expected to do the integral with a single substitution $w=\sin(x^2)$, instead of two substitutions. The reason I think so is that the question bundled the $x$ and the $\cos(x^2)$ together, in one convenient package which is almost the derivative of $\sin(x^2)$. So the shape in which the integrand was given was in itself a strong hint. If you had been asked the equivalent $$\int x (\sin(x^2))^5 \cos(x^2)\,dx,$$ the substitution $w=\sin(x^2)$ would be less obvious to the eye. Then $u=x^2$ would have been an even more natural first step.

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$x^2$ is "ugly". So we call it $u$, the first letter of "ugly". – Michael Lugo Jul 5 '11 at 18:21
@Michael Lugo: Perfect! Warning: I will steal it. – André Nicolas Jul 5 '11 at 19:12
It's not the most efficient way, but it is (IMO) the right way to show it to a novice. – Gerry Myerson Jul 6 '11 at 4:57
@Gerry Myerson: Of course all of us can write down the integral at once. But someone who asks the question needs a little more! – André Nicolas Jul 6 '11 at 6:34

Put $x^{2}=t$ then the integral becomes $$\frac{1}{2} \int \sin{x} \cos{x} \ dx$$ then put $\sin{x} =\nu$ to get the answer.

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Why are we making two substitutions?? – The Chaz 2.0 Jul 5 '11 at 19:12
Because not all of us can see our way clear through to the answer from the start. In fact, that's an important pedagogical point; I always tell my students they don't have to see how to do the problem, they just have to see how to make some progress. – Gerry Myerson Jul 6 '11 at 4:56

Hint: $2x\cos x^2$ is the derivative of...

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