Answer
\(
\sin x-\frac{\sin ^{3} x}{3}+C
\)
Explanation
(1) Use Pythagorean Identities: \(\cos ^{2} x=1-\sin ^{2} x\).
\[
\int\left(1-\sin ^{2} x\right) \cos x d x
\]
(2) Use Integration by Substitution.
Let \(u=\sin x, d u=\cos x d x\)
(3) Using \(u\) and \(d u\) above, rewrite \(\int\left(1-\sin ^{2} x\right) \cos x d x\).
\[
\int 1-u^{2} d u
\]
(4) Use Power Rule: \(\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\).
\[
u-\frac{u^{3}}{3}
\]
(5) Substitute \(u=\sin x\) back into the original integral.
\(\sin x-\frac{\sin ^{3} x}{3}\)
(6) Add constant.
\[
\sin x-\frac{\sin ^{3} x}{3}+C
\]