= Solution
Write $I_n=\int\sin^nx\,dx$. Apply <integration by parts> with $\sin^{n-1}x$ and $\sin x\,dx$:
$$
I_n=-\cos x\sin^{n-1}x+(n-1)\int\sin^{n-2}x\cos^2x\,dx.
$$
Using $\cos^2x=1-\sin^2x$ on the right gives $I_n=-\cos x\sin^{n-1}x+(n-1)(I_{n-2}-I_n)$. Rearranging proves
$$
\boxed{I_n=-\frac{\cos x\sin^{n-1}x}{n}+\frac{n-1}{n}I_{n-2}+C}.
$$
For $J_n=\int\cos^nx\,dx$, the analogous <integration by parts> uses $\cos^{n-1}x$ and $\cos x\,dx$, giving $J_n=\sin x\cos^{n-1}x+(n-1)(J_{n-2}-J_n)$. Hence
$$
\boxed{J_n=\frac{\sin x\cos^{n-1}x}{n}+\frac{n-1}{n}J_{n-2}+C}.
$$
An arbitrary <constant of integration> absorbs the choices in the lower-order <antiderivatives>.
Starting from $I_0=J_0=x$, successive use of $n=2,4,6$ gives
$$
\boxed{\int\sin^6x\,dx=
\frac{5x}{16}-\frac{\cos x\sin^5x}{6}
-\frac{5\cos x\sin^3x}{24}-\frac{5\sin x\cos x}{16}+C},
$$
$$
\boxed{\int\cos^6x\,dx=
\frac{5x}{16}+\frac{\sin x\cos^5x}{6}
+\frac{5\sin x\cos^3x}{24}+\frac{5\sin x\cos x}{16}+C}.
$$
These expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.
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