Q. 31

# Evaluate

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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