Double substitution in an indefinite integral

Question

I am training my self in solving integrals by substitution and I have a doubt in applying a double substitution.

To solve the integral $\int \sin^4{(7x)}\cos{(7x)}\, dx$ I have thought to do the substitution $t\to u$, so: $u=7x$ and then $du=7dx$, and hence: $$\int \sin^4{(7x)}\cos{(7x)}\, dx=\frac{1}{7}\int \sin^4{(u)}\cos{(u)}\, du$$ Now if I want to write $w=\sin{u}\implies dw=\cos{u}\,du$, so the integral becomes: $$\frac{1}{7}\int w^4 \, dw$$ $\textbf{Question:}$ is it right to write $dw=\cos{u}\,du$ and so considering that the derivative of $w$ with respect $u$ is $\cos{u}$ or I have to apply the chain rule of derivation and taking into the account that $u$ is a function of $x$ ($dw=7\cos{7x}\,dx$)?

Answer 1

No, it is right.

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Consider this: if you were given to solve $\int\sin^4u\cos u\,\text du$, then you would have done the same thing. It doesn't matter because now you are integrating something in $u$, not in $x$.

And even if you use chain rule, you will be getting $\text dx$, which you have to again change in order to make use of the $w$-substitution.

Hope this helps. Ask anything if not clear :)

Answer 2

There is no problem in doing more than one substitution. And there is no need to apply the chain rule here. At the end, you got $\frac1{28}w^5$. Since $w=\sin u$, this is the same thing as $\frac1{28}\sin^5u$. And, since $u=7x$, this is the same thing as $\frac1{28}\sin^5(7x)$.

Of course, after noticing that the substitutions $u=7x$ and $w=\sin u$ solve your problem, you could use the single substitution $w=\sin(7x)$ and then, yes, you should use the chain rule to know that then $\mathrm dw=7\cos(7x)\,\mathrm dx$.

Answer 3

You would see the answer to your question if instead you employed a single substitution of the form $$v = \sin 7x, \quad dv = 7 \cos 7x \, dx,$$ thus $$\int \sin^4 7x \cos 7x \, dx = \frac{1}{7} \int (\sin 7x)^4 (7 \cos 7x) \, dx = \frac{1}{7}\int v^4 \, dv.$$

In doing so, we employed the chain rule. Since $u(x) = 7x$ and $w(u) = \sin u$, we have $v(x) = w(u(x))$, and $$\frac{dv}{dx} = \frac{dw}{du}\cdot \frac{du}{dx} = (\cos u(x))(7) = 7 \cos 7x.$$