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Question

Evaluate π/20cos9xcos3x+sin3xdx.

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Solution

I=π20cos9xcos3x+sin3xdx(i)I=π20cos9(π2x)cos3(π2x)+sin3(π2x)I=π20sin9(x)sin3(x)+cos3(x)dx
adding (i) & (ii)
2I=π20cos9x+sin9xcos3x+sin3xdx2I=π20(cos3x)3+(sin3x)3sin3x+cos3xdx2I=(cos3x+sin3x)+(cos3x)2+(sin3x)sin3x.cos3xsin3x+cos3xdx2I=π20(cos3x)2+(sin3x)2sin3x.cos3xdx=π20cos3x{cos3xsin3x}+sin3x{sin3xcos3x}+{sin3x+cos3x}dx=π20cos3x{(cosx+sinx){1+cosx.sinx}}

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