Evaluate the definite integral of sin^2(x) cos^4(x) dx from x = pi/8 to x = pi/12.

Step-by-step solution

1. 1

   Notice that the limits run backwards. The lower limit $\tfrac{\pi}{8}=0.3927$ is larger than the upper limit $\tfrac{\pi}{12}=0.2618$. Since $\sin^2x\cos^4x\ge 0$ everywhere, the answer is forced to be negative:

   $\int_{\pi/8}^{\pi/12}=-\int_{\pi/12}^{\pi/8}$

   Deciding the sign in advance is the cheapest available check on the final number.

2. 2

   Reduce the powers with double-angle identities. Both exponents are even, so power reduction (not a $u$-substitution) is the right tool:

   $\sin^2x\cos^4x=\frac{1-\cos 2x}{2}\cdot\left(\frac{1+\cos 2x}{2}\right)^2=\frac{(1+\cos 2x)\left(1-\cos^2 2x\right)}{8}=\frac{\sin^2 2x+\sin^2 2x\cos 2x}{8}$

   The split is deliberate: the first piece needs another power reduction, the second is a perfect $u$-substitution.

3. 3

   Find an antiderivative term by term. Using $\sin^2 2x=\tfrac{1-\cos 4x}{2}$ for the first piece and $u=\sin 2x$ for the second,

   $\int\sin^2 2x\,dx=\frac{x}{2}-\frac{\sin 4x}{8},\qquad \int \sin^2 2x\cos 2x\,dx=\frac{\sin^3 2x}{6}$

   so

   $F(x)=\frac{x}{16}-\frac{\sin 4x}{64}+\frac{\sin^3 2x}{48}$

4. 4

   Evaluate at the two limits. At $x=\tfrac{\pi}{12}$: $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ and $\sin\tfrac{\pi}{6}=\tfrac12$, so

   $F\!\left(\tfrac{\pi}{12}\right)=\frac{\pi}{192}-\frac{\sqrt3}{128}+\frac{1}{384}$

   At $x=\tfrac{\pi}{8}$: $\sin\tfrac{\pi}{2}=1$ and $\sin\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}$, so $\sin^3\tfrac{\pi}{4}=\tfrac{\sqrt2}{4}$ and

   $F\!\left(\tfrac{\pi}{8}\right)=\frac{\pi}{128}-\frac{1}{64}+\frac{\sqrt2}{192}$

5. 5

   Subtract and put everything over 384.

   $F\!\left(\tfrac{\pi}{12}\right)-F\!\left(\tfrac{\pi}{8}\right)=-\frac{\pi}{384}+\frac{7}{384}-\frac{3\sqrt3}{384}-\frac{2\sqrt2}{384}=\frac{7-\pi-3\sqrt3-2\sqrt2}{384}$

   Numerically $7-3.14159-5.19615-2.82843=-4.16617$, and dividing by $384$ gives $\approx -0.010849$ — negative, as the reversed limits demanded. Direct numerical quadrature of the original integrand over the same reversed interval returns $-0.0108494$ $\checkmark$.