Is it possible to do integration with degrees rather than radians with trigonometric functions?
For example, if I want to find $\int_{0}^{180} \sin(x) \,dx $ where $x$ is measured in degrees, I can write it as $\int_{0}^{180} \sin(\frac{x\pi}{180})\,dx $ where $x$ is measured in radians, however this gives me $\frac{360}{\pi}.$ And if I wanted to differentiate $\sin(x)$ where $x$ is in degrees, I can get $\frac{\pi}{180} \cos(x).$ To get the correct answer for the integral, which is $2$, I have to cancel $\frac{180}{\pi}$ down to 1; why?
Best Answer
For the integral $$ \int_0^\pi\sin(x)\,dx $$ you want to change variables according to $x = \frac\pi{180}x^\circ$. You can't just change $x$, you also have to change $dx$, which tells you "how big" a change in $x$ is. An increment of 1 degree is not the same as an increment of 1 radian, we need to have $dx = \frac\pi{180}dx^\circ$. This means $$ \int_0^\pi\sin(x)\,dx = \int_0^{180}\sin(\pi x^\circ/180)\frac\pi{180}dx^\circ = \frac\pi{180}\int_0^{180}\sin(\pi x^\circ/180)\,dx^\circ. $$ This is a different integral from what you wrote: $$ \int_0^{180}\sin(\pi x^\circ/180)\,dx^\circ = \frac{180}\pi\int_0^\pi\sin(x)\,dx. $$