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

Question

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

Answer 1

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

Answer 2

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

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

Answer 3

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.

Answer 4

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.