Answer :

In this question, first of all we expand cot4x as

cot4x = (cosec2x – 1)2


= cosec4x – 2cosec2x + 1 …(1)


Now, write cosec4x as


cosec4x = cosec2xcosec2x


= cosec2x(1 + cot2x)


= cosec2x + cosec2xcot2x


Putting the value of cosec4x in eq(1)


cot4x = cosec2x + cosec2xcot2x – 2cosec2x + 1


= cosec2xcot2x – cosec2x + 1


y = ∫ cot4x dx


= ∫ cosec2x cot2x dx + ∫ - cosec2x + 1 dx


A = ∫cosec2x cot2x dx


Let, cot x = t


Differentiating both side with respect to x



-dt = cosec2x dx


= ∫-t2 dt


Using formula



Again, put t = cot x



Now, B= ∫-cosec2 x +1 dx


Using formula ∫cosec2 x dx= -cot x and ∫c dx=cx


B = cot x + x + c2


Now, the complete solution is


y = A + B



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