When to use Cohen's d and when t-test? When do I use Cohen's and when do I t-test? Probably in addition: What is the (conceptual) difference between them? Both tests are meant to study the difference between two distributions.
I just roughly know that Cohen's is used to calculate the effect size while t-test is meant to study whether there is a general difference between two distributions(?).
Formulas:
t-test: $ T = \frac{\bar{x} - \bar{y}}{ \sqrt{\frac{s_x^2}{n_x} + \frac{s_y^2}{n_y}} }  $
with $ s $ as a variance.
Cohen's $d = \frac{\mu_1 - \mu_2}{ \sigma } $
I see there are differences but it also looks very similar. Isn't it?

 A: Cohen's d seeks to tell you how big the standardized difference is between the two distributions.  It's very popular in areas like psychology where I think there are no obvious units you can use to describe the difference. In medical stats, I could say (for example) that your HbA1c levels were on average 5mg different in the two groups, and wouldn't need to use Cohen's d.
The t-test is an attempt to tell you have enough evidence to reject the idea that the difference is non-zero.   However, a non-zero difference could be, in practical terms, completely irrelevant.   Also, don't forget you have to make technical assumptions when using the t-test, e.g. you default to assume the two groups have the same variance. 
There are arguments that it is more useful to compare confidence or credible intervals estimated from the two samples.
There's an interesting article here: https://bmcresnotes.biomedcentral.com/articles/10.1186/s13104-015-1020-4
A: T-test is in complimentary relation with Cohen's $d$ (and equivalence tests using Cohen's $d$). 
T-test gives a p-value which is the probability of committing a Type I error. One can reject the null hypothesis, if the p-value is too small, but one cannot claim that the null hypothesis is true on the basis of p-value only, without risking of making a Type II error.
To assess the risk of Type II error one has to perform power testing, i.e. calculating the probability that the alternative hypothesis is correct. However, in the case of testing $H_0 : \mu =0$ against $H_1 : \mu \neq 0$ direct power calculation is impossible. 
One common solution to this problem is assuming the minimal size of the effect (here is where Cohen's $d$ comes in) and proving that the actual effect is smaller than this minimal size. This is known as "equivalence testing" or TOST (two one-sided tests). Here is a useful reference: https://www.ncbi.nlm.nih.gov/pubmed/28736600
There are alternative approaches, e.g., based on the use of Bayes factors. But these would take us too far from the core of your question.
