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Question

π/40sinx+cosxcos2x+sin4x

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Solution

I=x/40sinx+cosxcos2x+sin4xdx
=x/404(sinx+cosx)2.2cos2x+(2sin2x)2dx
=4x/40sinx+cosx2(1+cosx2)(1cosx2)2dx
=4x/40sinx+cosx3+cos22x=4x/40sinx+cosx4sin22xdx
Let sinxcosx=z(cosx+sinx)dx=dz
and z2=(sinxcosx)2z2=1sin2x
Or, sin2x=1z2
I=4x/40sinx+cosx4sin22xdx
=401dz22(1z2)2=014dz(3z2)(z2+1)
=01[1z2+1+13z2]dz
=[tan1z+123log3+z3z]01
=0[tan1(1)+123log313+1]
=π4+123log3+131
=π4+13log3+12Ans


1399084_1124654_ans_2521eccf6dd8414da7c414a81901d87b.png

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