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Question
secondary 4 | A Maths
2 Answers Below
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help integrate this pls
Date Posted: 10 months ago
Views: 163
∫(tan²x - x) dx
= ∫(sec²x - 1 - x) dx
(Recall that 1 + tan²x = sec²x so we can rewrite)
= tanx - x - x²/2 + c
or
tanx - x - ½x² + c
= ∫(sec²x - 1 - x) dx
(Recall that 1 + tan²x = sec²x so we can rewrite)
= tanx - x - x²/2 + c
or
tanx - x - ½x² + c
See 2 Answers
done
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clear
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Hope this helps. Regards
Date Posted:
10 months ago
helps lots , thanks !!!!
Great!
helpful !! thank u <3
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