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svlad2 [7]
3 years ago
5

Integrate sin²2x cos²2x dx

Mathematics
1 answer:
DochEvi [55]3 years ago
7 0
\sin^2x=\dfrac{1-\cos2x}2
\cos^2x=\dfrac{1+\cos2x}2

From the identities above, you have

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

Applying once more, you have

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

So,

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