Solving Riemann-Stieltjes integral:$\int_{- \pi/4}^{\pi/4} f(x)dg(x)$

Question

I'm having trouble solving this Riemann-Stieltjes integral:

$$\int_{- \pi/4}^{\pi/4} f(x)dg(x),$$ where $$f(x):= \begin{cases} \frac{\sin^4x}{\cos^2x}{} &\text{if }x\ge0, \\{}\\ \frac1{\cos^3x} &\text{if }x<0,\end{cases}$$

and $$g(x)=\begin{cases} \phantom{-} 1+\sin(x) &\text{if }-\pi/4 <x<\pi/4, \\ -1 &\text{otherwise}.\end{cases}$$

I believe the only jump discontinuities are at $-\pi/4$ and $\pi/4$. Which $g=-1$ at both of those points. I'm struggling with the rest. What formula should I be using to compute the integral and what should my answer look like? Thanks for any help!

Answer 1

Since $g$ is differentiable there is no trouble with sum this boils down to Riemann integral of $g'\cdot f$

Namely from here we have $$\color{blue}{\int_a^bf(x)dg(x)=\int_a^bf(x)g'(x)dx}$$ Therefore, $$\int_{-\pi/4}^{\pi/4} f(x) dg(x) = \int_{-\pi/4}^0 \frac 1{\cos^3 x} d(1+\sin x) + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(1+\sin x)\\=\int_{-\pi/4}^0 \frac 1{\cos^2 x} dx + \int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(\sin x)$$

But $$\int_{-\pi/4}^0 \frac 1{\cos^2 x} dx =\int^{\pi/4}_0 (\tan x)' dx =1$$

and $$\int_0^{\pi/4} \frac{\sin^4 x}{\cos^2 x} d(\sin x) = \int_0^{\pi/4} \frac{\sin^4 x}{1-\sin^2 x} d(\sin x) =\int_0^{\sqrt{2}/2} \frac{t^4 }{1-t^2 } dt\\=\int_0^{\sqrt{2}/2} -t^2+\frac{t^2 }{1-t^2 } dt$$ $$ $$