= Solution
The product in the integrand simplifies to $\sin x\cos^3x$ below its upper endpoint. With the <integration by substitution> $t=\cos x$, the limits become $c$ and zero and $dt=-\sin x\,dx$. Therefore
$$
I=-\int_c^0t^3\,dt=\int_0^ct^3\,dt=J,
\qquad\boxed{I=J=\frac{c^4}{4}.}
$$
Although the <tangent function> itself is undefined at the upper endpoint, the product has a zero <limit> there, so its continuous extension, or its <improper integral>, gives this same result.
To obtain $J$ directly from a <Riemann sum>, divide $[0,c]$ into $n$ equal pieces of width $c/n$ and use right endpoints $t_k=kc/n$. The sum is
$$
\sum_{k=1}^n\left(\frac{kc}{n}\right)^3\frac cn
=\frac{c^4}{n^4}\frac{n^2(n+1)^2}{4}
=\frac{c^4}{4}\left(1+\frac1n\right)^2.
$$
The integrand is continuous, so the <limit> of these <Riemann sums> is its <definite integral>, namely $c^4/4$, agreeing with the <antiderivative> calculation.
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