Trapezoidal rule to find $\int_{0}^{1} \frac {\cos(2x)}{x^{1/3}} dx$ I am supposed to find the integration of the given function in the interval $[0,1]$ using trapezoidal rule _(as an assignment problem)_. $$ \frac {\cos(2x)}{x^\frac{1}{3}} $$ As it can be seen, the value of the function is not defined at $x=0$ and hence, it seems impossible to find the solution using trapezoidal rule. Am I thinking the right way, or there exists some method to find the solution?

The behavior of the singularity at the origin is like $\dfrac1{x^{1/3}}$. It is advisable to remove it and integrate it separately:

$$\int_0^1\frac {\cos(2x)}{x^{1/3}}dx=\int_0^1\frac{\cos(2x)-1}{x^{1/3}}dx+\int_0^1\frac{dx}{x^{1/3}}=-2\int_0^1\frac{\sin^2(x)}{x^{1/3}}dx+\frac32x^{2/3}\Big|_0^1.$$

Unfortunately, this form is still unsuitable for trapezoidal interpolation, as the second derivative has a $x^{-1/3}$ factor, making the remainder potentially large (magenta curve).

Alternatively, use the change of variable $x=t^\alpha$, to turn the integral to

$$\alpha\int_0^1\cos(2t^\alpha) t^{\alpha-1-\alpha/3}\,dt.$$

You have the option to choose $\alpha=\frac32$ to get rid of the power factor completely

$$\frac32\int_0^1\cos(2t^{3/2})\,dt,$$ or $\alpha=3$ to avoid fractional exponents $$2\int_0^1\cos(2t^3)t\,dt.$$

In both cases, the second derivative is well-behaved. You should prefer $\alpha=\frac32$ (blue curve).

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