I am trying to prove that by Mathematical Induction :
$$\sin(x)= \int_{0}^{x} \cos(t) dt $$ $$= x cos(x)+\frac{x^{2}}{2} \sin(x)- \frac{x^{3}}{3!} \cos(x)- \frac{x^{4}}{4!} \sin(x)+\frac{x^{5}}{5!} \sin(x).........$$ $$ = cos(x) \bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}....\bigg)+sin(x) \bigg(x-\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}....\bigg)$$ $$ = cos(x) \bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}....\bigg)-sin(x) \bigg(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}....\bigg)+sin(x)$$
So what I have done so far, that I have integrated by parts several time, but what I do not understand how induction work on integration, like what is my base case and what is my inductive step.
$$\int_{0}^{x} \cos(t) dt $$ Let $$u= cos(t)$$ and $$du = -sin(t) dt $$ Then, let $$ dv= dt$$ and $$v=t$$
Then we have, $$\int_{0}^{x} \cos(t) dt = t cos(t) \mid_{0}^{x}+ \int_{0}^{x} tsin(t) dt$$ $$= x cos(x) + \int_{0}^{x} tsin(t) dt$$
Then I consider $$ \int_{0}^{x} tsin(t) dt$$ I let $$ s=sin(t)$$ and $$ ds=cos(t) dt$$ Then $$dr=t$$ $$ r=\frac{t^2}{2}$$
So, $$\int_{0}^{x} tsin(t) dt= \frac{t^2sin(t)}{2} \mid_{0}^{x} - \frac{1}{2} \int_{0}^{x} t^{2}cos(t) dt$$ Which is $$\int_{0}^{x} tsin(t) dt= \frac{x^2sin(x)}{2} - \frac{1}{2} \int_{0}^{x} t^{2}cos(t) dt$$