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Question:

The integral ∫sin2xcos2x(sin5x+cos3xsin2x+sin3xcos2x+cos5x)²dx is equal to

⅓+cot3x+C

⅓(1+tan3x)+C

1/(1+cot3x)+C

⅓(1+tan3x)+C

Solution:

I=∫sin2xcos2xdx[sin2x(sin3x+cos3x)+cos2x(sin3+cos3x)]²=∫sin2xcos2xdx[(sin2x+cos2x)(sin3x+cos3x)]²=∫sin2xcos2xdx(sin3x+cos3x)²=∫sin2xcos2xdxcos⁶x(tan3x+1)²=∫tan2xsec2x(tan3x+1)²dxPut tan3x+1=t ∴ 3tan2xsec2xdx=dt=∫dt/(3t²)=⅓t+1=⅓(tan3x+1)+c