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Question

Evaluate the following integrals:

1sin4x+sin2x cos2x+cos4xdx

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Solution

Let I=1sin4x+sin2x cos2x+cos4xdx =1sin2x+cos2x2-sin2x cos2xdx =11-sin2x cos2xdx =1cos4x1cos4x-sin2xcos2xdx =sec2x1+tan2xsec4x-tan2xdx Let tanx=t On differentiating both sides, we get sec2x dx=dtI=1+t21+t22-t2dt =1+t2t4+t2+1dt =1t2+1t2+1+1t2dt =1t2+1t-1t2+3dt Let t-1t=u On differentiating both sides, we get 1+1t2 dt=duI=1u2+3du =13tan-1u3+c =13tan-1t-1t3+c =13tan-1tanx-cotx3+cHence, 1sin4x+sin2x cos2x+cos4xdx=13tan-1tanx-cotx3+c

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