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secondary 4 | A Maths
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Good afternoon, Please help me w/ this. Thank you for your effort and help:)
Date Posted: 5 years ago
Views: 210
Ln(value)
When we see ln we put the value at the denominator and put its differentiated value at the top
Hence for ln(cos x)
Differentiation of cos x is - sin x
Therefore it is
-sinx/cosx = - tanx
When we see ln we put the value at the denominator and put its differentiated value at the top
Hence for ln(cos x)
Differentiation of cos x is - sin x
Therefore it is
-sinx/cosx = - tanx
We know integration is the converse of differentiation.
And from the first part we see that differentiating ln(cos x) gives us - tan x. Therefore we know by integrating - tanx we have ln(cosx)
However, now we are presented with tan x only. Therefore to solve that we put two minus sign. One inside of the integral sign and one outside. This way we can be integrating - tanx.
So we get - intégral(-tan x).
And we get - ln(cos x)
The computation can then be done by - ln(cos pi/3) + ln(cos 0)
And from the first part we see that differentiating ln(cos x) gives us - tan x. Therefore we know by integrating - tanx we have ln(cosx)
However, now we are presented with tan x only. Therefore to solve that we put two minus sign. One inside of the integral sign and one outside. This way we can be integrating - tanx.
So we get - intégral(-tan x).
And we get - ln(cos x)
The computation can then be done by - ln(cos pi/3) + ln(cos 0)
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Good afternoon Sonia! Here are my workings for the question.
Date Posted:
5 years ago
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