Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
pls help need it urgently
Date Posted: 3 years ago
Views: 471
INT (sin x - cos x)^2 dx
= INT (sin^2 x - 2 sin x cos x + cos^2 x) dx
= INT (1 - 2 sin x cos x) dx
= INT (1 - sin 2x) dx
= x + 0.5 cos 2x + constant
= x + 0.5 (1 - 2 sin^2 x) + constant
= x + 0.5 - sin^2 x + constant
= x - sin^2 x + another constant
= INT (sin^2 x - 2 sin x cos x + cos^2 x) dx
= INT (1 - 2 sin x cos x) dx
= INT (1 - sin 2x) dx
= x + 0.5 cos 2x + constant
= x + 0.5 (1 - 2 sin^2 x) + constant
= x + 0.5 - sin^2 x + constant
= x - sin^2 x + another constant
Alternatively,
INT (1 - 2 sin x cos x) dx
= x - sin^2 x + constant
Because sin x cos x is of the form [f(x)]^n times f’(x) where f(x) = sin x, n = 1 and f’(x) = cos x; this integrates to [f(x)] ^ (n + 1), whole thing divided by (n + 1)
But this technique requires a great deal of observation and it’s more suited for A Levels.
INT (1 - 2 sin x cos x) dx
= x - sin^2 x + constant
Because sin x cos x is of the form [f(x)]^n times f’(x) where f(x) = sin x, n = 1 and f’(x) = cos x; this integrates to [f(x)] ^ (n + 1), whole thing divided by (n + 1)
But this technique requires a great deal of observation and it’s more suited for A Levels.
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Please check
Date Posted:
3 years ago
360 Tutoring Program