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secondary 4 | A Maths
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Hi, I don't know how to do 10(ii), I know how to solve it individually but the hence part idk how to relate part (i) and (ii)
Date Posted: 4 years ago
Views: 193
d/dx (2x + x sin x)
= 2 + sinx + x cos x
= 2 + sinx + x cos x
So from i) we know that :
d/dx (2x + x sin x) = 2 + sinx + x cos x
Then ,
2 d/dx (2x + x sin x) = 4 + 2 sin x + 2x cos x
2x cos x = 2 d/dx (2x + x sin x) - 4 - 2 sin x
Next ,
∫ 2x cos x dx = ∫ ( 2 d/dx (2x + x sin x) - 4 - 2 sin x) dx
∫ 2x cos x dx = 2 ∫ d/dx(2x + x sin x) dx - ∫ 4 dx - 2 ∫ sin x dx
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] - 2[-cos x]
(Integrating a derivative of an expression is like working backwards, it just gets you back to the expression itself)
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] + 2[cos x]
Sub in your lower and upper limits accordingly
d/dx (2x + x sin x) = 2 + sinx + x cos x
Then ,
2 d/dx (2x + x sin x) = 4 + 2 sin x + 2x cos x
2x cos x = 2 d/dx (2x + x sin x) - 4 - 2 sin x
Next ,
∫ 2x cos x dx = ∫ ( 2 d/dx (2x + x sin x) - 4 - 2 sin x) dx
∫ 2x cos x dx = 2 ∫ d/dx(2x + x sin x) dx - ∫ 4 dx - 2 ∫ sin x dx
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] - 2[-cos x]
(Integrating a derivative of an expression is like working backwards, it just gets you back to the expression itself)
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] + 2[cos x]
Sub in your lower and upper limits accordingly
See the previous comment
Edited that comment , fixed a few errors
For the last statement,
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] + 2[cos x] can be further simplified :
∫ 2x cos x dx = [4x + 2x sin x - 4x + 2 cos x]
∫ 2x cos x dx = [2x sin x + 2 cos x]
Sub in your lower and upper limits accordingly as before
∫ 2x cos x dx = 2 [2x + x sin x] - [4x] + 2[cos x] can be further simplified :
∫ 2x cos x dx = [4x + 2x sin x - 4x + 2 cos x]
∫ 2x cos x dx = [2x sin x + 2 cos x]
Sub in your lower and upper limits accordingly as before
Hang on, writing this up
Actually he's already seen the answer after posting his previous comment (which he deleted).
See 2 Answers
Missing a brackets for the first expression
i.e ∫(2 + sinx) dx is the correct presentation
i.e ∫(2 + sinx) dx is the correct presentation
done
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clear
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Two approaches to the question, both are equally valid and can be used in the O Levels. You may find the first method easier, but personally I always use the second method.
Date Posted:
4 years ago
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