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Worked solution

Turn a same-angle product into one sine

Find ∫sin⁡3xcos⁡3x dx\int\sin3x\cos3x\,dx.

CJC · Lecture · Example 14

Use double angle

sin⁡3xcos⁡3x=12sin⁡6x\sin3x\cos3x=\frac12\sin6x. Therefore ∫sin⁡3xcos⁡3x dx=12∫sin⁡6x dx=12(−cos⁡6x/6)+C=−112cos⁡6x+C\int\sin3x\cos3x\,dx=\frac12\int\sin6x\,dx=\frac12(-\cos6x/6)+C=-\frac1{12}\cos6x+C.

How did this part go?

Common mistakes

Include both the half from the identity and the sixth from the inner derivative.

Read the hint