We can solve this problem by simplifying using trigonometric identities but, I was trying to find different methods to solve trigonometric integration problems, and Taylor series is something which I am trying to use to solve some.
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}…$$
So we have $$\int (\sin x)^4= \int (x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}…)^4$$
But I don’t know how to proceed.
Also is it advisable to use this method for such problems, or am I overcomplicating it?
Hint:
When you can directly find using substitutions or some other method you need not use the Taylor series,
Observe that $$ \int \sin^4 x\, dx = \int \sin^2x \,dx - \int \sin^2x \cos^2 x \, dx =\int \sin^2x \,dx - \frac14\int \sin^2{2x} \, dx $$
and you can easily evaluate $\int \sin^2 t \, dt$ by writing $\sin^2t $ in terms of $\cos{2t}$.
Update:
Let $$ S = \sum_{k=0}^{\infty} a_k x^k$$ we refer here, $$S^2 = \sum_{k=0}^{\infty} \sum_{j=0}^{k}a_ja_{k-j}x^k$$
I think directly considering $\sin^4(x)$ would be laborious to compute coefficients as we have to square twice, instead that I suggest you to write and calculate square using the above equation $$ \sin^2 x = \frac{1 - \cos{2x}}{2} = \sum_{k=1}^{\infty} \frac{ 2^{2k-1}x^{2k} (-1)^k}{(2k)!} $$
This may help you in guessing a simple form for the coefficients after the integration thereby getting a simple form for the antiderivative.