Question

∫[(sec²x – 2009)/(sin²⁰⁰⁹x)]dx = __________ + c

(a) [(cotx)/(sin²⁰⁰⁹x)]

(b) [(– cotx)/(sin²⁰⁰⁹x)]

(c) [(tanx)/(sin²⁰⁰⁹x)]

(d) [(– tanx)/(sin²⁰⁰⁹x)]

Answer 1

The correct option (c) [(tanx)/(sin²⁰⁰⁹x)]

Explanation:

I = ∫[(sec²x – 2009)/(sin²⁰⁰⁹x)]

multiplying by sin²⁰⁰⁹x

I = ∫[(sin²⁰⁰⁹x ∙ sec²x – 2009 ∙ sin²⁰⁰⁹x)/(sin²⁰⁰⁹x)²]

= ∫[(sin²⁰⁰⁹x ∙ sec²x – {(2009 ∙ sin²⁰⁰⁹x ∙ tanx)/(tanx)})/(sin²⁰⁰⁹x)²]dx

= ∫[(sin²⁰⁰⁹x ∙ sec²x – 2009 ∙ sin²⁰⁰⁸x ∙ cosx ∙ tanx)/(sin²⁰⁰⁹x)²]dx

= ∫[{sin²⁰⁰⁹x ∙ (d/dx)tanx – tanx ∙ (d/dx)(sin²⁰⁰⁹x)dx}/(sin²⁰⁰⁹x)²]

= ∫(d/dx)[(tanx)/(sin²⁰⁰⁹x)]dx

= [(tanx)/(sin²⁰⁰⁹x)] + c