Evaluate the integral: I = integral sin^4x cos^4x dx

Evaluate the integral: $$ I = \int \sin^4x\,\cos^4x\,dx. $$

Step 1 · Combine via the double-angle identity

Since $\sin^4x\cos^4x = (\sin x\cos x)^4$ and $\sin x\cos x = \tfrac12\sin2x$:

$$ \sin^4x\cos^4x = \left(\frac12\sin2x\right)^4 = \frac{1}{16}\sin^4(2x). $$

Step 2 · Reduce $\sin^4(2x)$

Using the same power-reduction steps as Problem 133 (with $u=2x$ in place of $x$), $\sin^4u=\tfrac38-\tfrac12\cos2u+\tfrac18\cos4u$. With $u=2x$:

$$ \sin^4(2x) = \frac38 - \frac12\cos4x + \frac18\cos8x. $$

Step 3 · Combine with the leading factor

Substituting into $\sin^4x\cos^4x=\tfrac1{16}\sin^4(2x)$:

$$ \sin^4x\cos^4x = \frac{1}{16}\left(\frac38 - \frac12\cos4x + \frac18\cos8x\right) = \frac{3}{128} - \frac{1}{32}\cos4x + \frac{1}{128}\cos8x. $$

Step 4 · Integrate term by term

$$ I = \int \left(\frac{3}{128} - \frac{1}{32}\cos4x + \frac{1}{128}\cos8x\right) dx = \frac{3}{128}x - \frac{1}{128}\sin4x + \frac{1}{1024}\sin8x + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• A common pitfall when combining $\sin^4x\cos^4x=\tfrac1{16}\sin^4(2x)$ with the power-reduction expansion of $\sin^4(2x)$ is to forget to distribute the leading $\tfrac1{16}$ across every term of the expansion — dropping it multiplies the whole final answer by 4.
• The result can be checked by differentiating $\tfrac{3}{128}x-\tfrac1{128}\sin4x+\tfrac1{1024}\sin8x$ and confirming it returns $\sin^4x\cos^4x$.