Evaluate the integral of sin^2(x) cos^3(x) with respect to x.

Step-by-step solution

1. 1

   Spot the odd power. Cosine appears to the odd power $3$. Whenever one of the two trig factors has an odd exponent, that factor can supply the differential for a substitution, so no reduction formula is required.

2. 2

   Peel off one cosine. Write $\cos^3 x=\cos^2 x\cdot\cos x$, so the integral becomes $\int\sin^2 x\,\cos^2 x\,\cos x\,dx$.

3. 3

   Convert the remaining even part. Using the Pythagorean identity $\cos^2 x=1-\sin^2 x$, the integrand is $\int\sin^2 x\left(1-\sin^2 x\right)\cos x\,dx$. Everything is now a function of $\sin x$ times $\cos x\,dx$.

4. 4

   Substitute $u=\sin x$. Then $du=\cos x\,dx$ and the integral collapses to the polynomial $\int u^2\left(1-u^2\right)du=\int\left(u^2-u^4\right)du$.

5. 5

   Integrate and back-substitute. $\frac{u^3}{3}-\frac{u^5}{5}+C=\frac{\sin^3 x}{3}-\frac{\sin^5 x}{5}+C$.

6. 6

   Check numerically. Evaluating this antiderivative between $0$ and $1$ gives $0.11423043$, and Simpson's rule applied to the original integrand over the same interval returns $0.11423043$ — the two agree.