My question is as follows:
Use induction to prove the following formula for $n \geq 2$
Solution (what I have tried so far):
Base case: $n=2$
RHS: $\int \sin^2xdx = \int 1/2 (1-\cos(2x))dx$
LHS: $\dfrac{-1}{2} \cos x \sin x + \dfrac{2-1}{2} \int dx$
RHS = LHS so base case holds (supposed to be but I haven't worked it out).
Induction Hypothesis (assume true for $n = k$): $\int \sin^kx dx = \dfrac{-1}{k} \cos x(\sin^{k-1}x) + \dfrac{k-1}{k} \int(\sin^{k-2}x)dx$
Induction Step: ($n = k + 1$) requires $\int (\sin^{k+1}x) dx = \dfrac{-1}{k+1} \cos x(\sin^kx) + \dfrac{k}{k+1} \int (\sin^{k-1}x)dx$
So, $\int (\sin^{k+1}x) dx = \in (\sin^kx) dx + \int \sin x dx$
Proof by induction is not necessary here. Suppose $n\geq 2$. First, think of $\sin^n(x)$ as a product $\sin^{n-1}(x)\cdot\sin(x)$. We can then use integration by parts: Take $f'=\sin(x)$ and $g=\sin^{n-1}(x)$ where the rule is $\int f'g=fg-\int fg'$.
This gives the following: $$ \begin{align*} \int\sin^n(x)dx &=-\cos(x)\sin^{n-1}(x)+(n-1)\int\cos^2(x)\sin^{n-2}(x)dx\\ &=-\cos(x)\sin^{n-1}(x)+(n-1)\int(1-\sin^2(x))\sin^{n-2}(x)dx\\ &=-\cos(x)\sin^{n-1}(x)+(n-1)\int\sin^{n-2}(x)dx-(n-1)\int\sin^{n}(x)dx\\ \Rightarrow n\int\sin^n(x)dx&=-\cos(x)\sin^{n-1}(x)+(n-1)\int\sin^{n-2}(x)dx\\ \Rightarrow\int\sin^n(x)dx&=-\frac{1}{n}\cos(x)\sin^{n-1}(x)+\frac{n-1}{n}\int\sin^{n-2}(x)dx \end{align*} $$ as required.