Question

Integrate sin²2x cos²2x dx

Answer 1

`\sin^2x=\frac{1-\cos2x}2`

`\cos^2x=\frac{1+\cos2x}2`

From the identities above, you have


`\sin^22x\cos^22x=\frac{(1-\cos4x)(1+\cos4x)}4=\frac{1-\cos^24x}4`

Applying once more, you have


`\frac{1-\cos^24x}4=\frac{1-\frac{1+\cos8x}2}4=\frac{1-\cos8x}8`

So,


`\displaystyle\int\sin^22x\cos^22x\,\mathrm dx=\frac18\int(1-\cos8x)\,\mathrm dx=\frac x8-\frac1{64}\sin8x+C`