15: There is a factorial of odd numbers in the denominator, which clearly points to the sine expansion. There is only one problem, the term 2k + 3 in the numerator.

The usual way of handling such a situation is to also create 2k + 3 in the denominator by integrating x^2k + 2, which is something that can be easily created there. We see that the procedure will involve integration, which is usually handled more easily if you first denote the sum of the given series as f (x) and then manipulate simultaneously both sides of this equality.

There is an alternative way to work with the 2k + 3 in the numerator. Try to think of some way to achieve cancelling with the factorial.

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