# Integrating $\left|f(x)\right|$ by pulling out $\mathrm{sgn}(f(x))$ from the integral

I tried doing the following integral: $$\int_{0}^{\pi/4}\sqrt{1-\sin2x}\mathrm dx$$. Firstly I completed the square by rewriting $$1$$ as $$\sin^2x+\cos^2x$$ to get the integral revised to this form: $$I=\int_{0}^{\pi/4}\left|\cos x-\sin x\right|\mathrm dx$$ Now clearly I can figure out which trig function is greater on that interval to get rid of that absolute value sign. But let's assume had it been an indefinite integral and I was to find the general antiderivative of the integrand.

I tried putting it into Integral Calculator which pulled out $$\mathrm{sgn}(\cos x-\sin x)$$ and simplified the integral as follows: $$\int \left|\cos x-\sin x\right|\mathrm dx=\mathrm{sgn}(\cos x-\sin x)\int(\cos x-\sin x)\mathrm dx$$

My query is: Can it be done for all integrands contained in absolute value expressions? Does it work for $$\int \left|x\right|\mathrm dx$$ or literally any other expression? I would love to get details as to why it works if it does, maybe in the form of a proof. Many thanks.

Edit: As Gerry Myerson pointed out that this formula: $$\int \left|f'(x)\right|\mathrm dx=f(x)\mathrm{sgn}(f(x))$$ does not always hold in the link shared by Yuito Cheng in the comments. I would like to follow up to the original question.

Under what conditions on $$f'(x)$$ is this formula valid?

• Related (if not a duplicate): What is the indefinite integral of |𝑓(𝑥)|? – YuiTo Cheng Apr 23 at 5:14