Understanding subsitution by parts of $\int_0^{1/4}x\sin^{-1}(4x)dx$ I am doing an Integration by Parts exercise and can't wrap my head around one of the answers:

Consider the definite integral
$$\int_0^{1/4}x\sin^{-1}(4x)dx$$

*

*We select $u = \sin^{-1}(4x)$.

*We select $dv = xdx$.

*Consider the known formula $$\int udv = uv - \int vdu$$

*Then, after integrating by parts, we obtain the integral $$\int_0^{1/4}vdu = \int_0^{1/4}f(x)dx$$ on the right hand side where
$$f(x) = \frac{x^2}{2} \cdot \frac{1}{(\sqrt{1-(4x)^2})} \cdot 4$$


I get it all up to this point, but then the exercise says this:



*The most appropriate subsitution to simplify this integral is $x = g(t)$ where $$g(t) = \frac{\sin(t)}{4}$$


*

*It's not clear to me what do they mean by "simplify this integral". What's there to simplify and why?

*And of course, why $\frac{\sin(t)}{4}$? I can almost see what's going on because that integral looks a lot like the differentiation of $\arcsin(x)$, but still failing to make the connection.

What is it that I'm missing? What was the train of thought behind that substitution?
 A: I think the suggestion is implying that it is "simpler" to deal with an integral that has $\sin(t)$ in it, as opposed to $\sin^{-1}(4x)$.
If you substitute $x=\frac{\sin(t)}{4}$, then you will be able to use
$$\sin^{-1}(4x) = \sin^{-1}(\sin t) = t.$$
After the substitution, you will still have to use integration by parts, and it might help to use a trig identity ($(\sin t)(\cos t) = \frac12 \sin 2t$), but arguably this is simpler than the two integration by parts solution that was posted earlier.
A: The rationale behind that substitution is that $\sqrt{1-(4x)^2}$ becomes something nicer - namely, $|\cos x|$, which is just $\cos x$ on the interval of integration.
I think it would have been clearer if they made that substitution at the beginning, rather than at the end. That way, you could see that $\sin^{-1}(4x)$ becomes $t$, and the original integral would become $$\frac{1}{16}\int_0^{\pi/2}t\sin(t)\cos(t)dx$$ which is easier to integrate in my opinion.
A: Integrate by parts twice as follows
\begin{align}
\int_0^{1/4}x\sin^{-1}(4x)dx 
&=\frac12 \int_0^{1/4}\sin^{-1}(4x)d(x^2)\\
&=\frac12x^2\sin^{-1}(4x)\bigg|_0^{1/4}-2\int_0^{1/4} \frac{x^2}{\sqrt{1-16x^2}}dx\\
&=\frac\pi{64}+\frac1{16}\int_0^{1/4} \frac{x}{\sqrt{1-16x^2}}d(1-16x^2) \\
&=\frac\pi{64}-\frac1{16}\int_0^{1/4} \frac{1}{\sqrt{1-16x^2}}dx\\
&=\frac\pi{128}
\end{align}
A: How do you deal with a problem like
$$\int\frac{\text{d}x}{1-x^2}$$
Most likely, you would substitute $x=\sin u$ so that you can use the trig identity for the denominator and everything simplifies.
In your regard, there is a $4x$ instead of $x$, so they just substituted $4x=\sin u$, (and used their fancy way to write the same thing!)
Hope this helps. Ask anything if not clear :)
