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?
Best Answer
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.