Question

Evaluate ∫ cosec²x cos²2x dx

Answer 1

∫ cosec²x cos²2x dx

= ∫cosec²x(cos²x-sin²x)²dx

Opening the square,

On multiplying (1-sin²x)(1-sin²x) equation reduces to, 

= ∫(cosec²x-2+sin²x-2cos²x+ sin²x)dx 

= ∫(cosec²x-2+2sin²x-2cos²x)dx 

= ∫(-2(cos²x-sin²x)+cosec²x-2)dx 

= ∫(-2cos2x+cosec²x-2)dx 

On solving this we get our answer i.e,

= \(\frac{-2sin2x}{2}\) - cotx - 2x + c

=-sin2x - cotx - 2x + c