Question

Integration of cos2x/cos^2x sin^2x = -secx cosecx

Answer 1

∫cos2x.dx/sin²x.cos²x

[cos2x = cos²x - sin²x , use this here ]
⇒∫(cos²x - sin²x ).dx/sin²x.cos²x
⇒∫dx/sin²x - ∫dx/cos²x 
⇒∫cosec²x.dx - ∫sec²x.dx 
⇒ -cotx - tanx 
⇒-(sinx/cosx + cosx/sinx)
⇒-(sin²x + cos²x )/sinx.cosx 
⇒-1/sinx.cosx
⇒ - secx.cosecx 

Answer 2

Answer:

Step-by-step explanation:

∫cos2x.dx/sin²x.cos²x

[cos2x = cos²x - sin²x , use this here ]

⇒∫(cos²x - sin²x ).dx/sin²x.cos²x

⇒∫dx/sin²x - ∫dx/cos²x 

⇒∫cosec²x.dx - ∫sec²x.dx 

⇒ -cotx - tanx 

⇒-(sinx/cosx + cosx/sinx)

⇒-(sin²x + cos²x )/sinx.cosx 

⇒-1/sinx.cosx

⇒ - secx.cosecx 

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