Mathematics: Post your doubts here!

Rutzaba

Rutzaba

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Ok u simplify it frst by taking out the conjugate and multiplying this as well as dividinh this by it
 
D0cEngi

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Can someone plz post the notes for Integration of the sin,cos and tan etc...(P3)
Thanks in advance...
 
VelaneDeBeaute

VelaneDeBeaute

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D0cEngi said:
Plz solve Q.5 of this ppr...: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_3.pdf
here is the ms: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_ms_3.pdf
Click to expand...

For the first part, you should know R cos ( θ - α ) = R cos θ cos α + R sin θ sin α
Now compare this with the given form = cos θ + √3 sin θ
R is usually calculated as the under root of squares of the sum of the coefficients of both cos and sin.
So, R = √[1^2 + (√3)^2 ] = √(1+3) = √4 = 2
α can then be found as tan inverse of (√3 / 1), which gives us the value of 60 degrees.

For the second part, substitute the denominator as 2 cos ( θ - 60 ). Then square it, which gives us the denominator as 4 cos^2 ( θ - 60 ). Take 1/4 out because it is the constant. Inverse of cos^2 is actually sec^2 ( θ-60) and the integration of sec^2 is tan ( θ-60). Hence put in the limits and you get the answer. :)
 
D0cEngi

D0cEngi

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VelaneDeBeaute said:
For the first part, you should know R cos ( θ - α ) = R cos θ cos α + R sin θ sin α
Now compare this with the given form = cos θ + √3 sin θ
R is usually calculated as the under root of squares of the sum of the coefficients of both cos and sin.
So, R = √[1^2 + (√3)^2 ] = √(1+3) = √4 = 2
α can then be found as tan inverse of (√3 / 1), which gives us the value of 60 degrees.

For the second part, substitute the denominator as 2 cos ( θ - 60 ). Then square it, which gives us the denominator as 4 cos^2 ( θ - 60 ). Take 1/4 out because it is the constant. Inverse of cos^2 is actually sec^2 ( θ-60) and the integration of sec^2 is tan ( θ-60). Hence put in the limits and you get the answer. :)
Click to expand...
Thank u so very much....It's awesome...:) (y)
 
D0cEngi

D0cEngi

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VelaneDeBeaute said:
For the first part, you should know R cos ( θ - α ) = R cos θ cos α + R sin θ sin α
Now compare this with the given form = cos θ + √3 sin θ
R is usually calculated as the under root of squares of the sum of the coefficients of both cos and sin.
So, R = √[1^2 + (√3)^2 ] = √(1+3) = √4 = 2
α can then be found as tan inverse of (√3 / 1), which gives us the value of 60 degrees.

For the second part, substitute the denominator as 2 cos ( θ - 60 ). Then square it, which gives us the denominator as 4 cos^2 ( θ - 60 ). Take 1/4 out because it is the constant. Inverse of cos^2 is actually sec^2 ( θ-60) and the integration of sec^2 is tan ( θ-60). Hence put in the limits and you get the answer. :)
Click to expand...
Can u plz also help me with Q.7 of the same ppr...? :D
 
VelaneDeBeaute

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D0cEngi said:
Can u plz also help me with Q.7 of the same ppr...? :D
Click to expand...
In Q.7. part 1, you need to pull up through 'integration by substitution method'.
First, see the given equation in question u =√x
Now differentiate u w.r.t. x
You'll get du/dx = 1 /(2√x)
dx = 2√x du
(Now substitute √x for u as indicated in the start.
You have dx = 2u du
Look into the question now. You put the dx = 2u du in place of dx written there, as well as substitute the x's in the denominator for u's and hence it becomes 1/u(4-u)
Last thing, we need to change the limits. A limit of 4 to 1 for x means that we had to solve for x=4 and x=1. Similarly, look into the equation u = √x . You put the values of x and you get the the limits of u. E.g. u = √4 ; u = 2.
Hence, expressed in the form as given in the question.
Part b is quite easy. All you need to do is to express in the partial fractions, integrate and put in the limits. :)
 
D0cEngi

D0cEngi

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VelaneDeBeaute said:
In Q.7. part 1, you need to pull up through 'integration by substitution method'.
First, see the given equation in question u =√x
Now differentiate u w.r.t. x
You'll get du/dx = 1 /(2√x)
dx = 2√x du
(Now substitute √x for u as indicated in the start.
You have dx = 2u du
Look into the question now. You put the dx = 2u du in place of dx written there, as well as substitute the x's in the denominator for u's and hence it becomes 1/u(4-u)
Last thing, we need to change the limits. A limit of 4 to 1 for x means that we had to solve for x=4 and x=1. Similarly, look into the equation u = √x . You put the values of x and you get the the limits of u. E.g. u = √4 ; u = 2.
Hence, expressed in the form as given in the question.
Part b is quite easy. All you need to do is to express in the partial fractions, integrate and put in the limits. :)
Click to expand...
Thank u..:)
 
J

Just visiting

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Hi guys, I need help..... in modulous functions
why x+|2x-1|=3
has solution -2 and 4/3
while 3+|2x-1|=x
has no solution.
but when i solve both with squaring side, i get same solutions
please give a clear explanation
 
J

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can some1 help in solving 16b and 17a,c,d
 

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D0cEngi

D0cEngi

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VelaneDeBeaute said:
In Q.7. part 1, you need to pull up through 'integration by substitution method'.
First, see the given equation in question u =√x
Now differentiate u w.r.t. x
You'll get du/dx = 1 /(2√x)
dx = 2√x du
(Now substitute √x for u as indicated in the start.
You have dx = 2u du
Look into the question now. You put the dx = 2u du in place of dx written there, as well as substitute the x's in the denominator for u's and hence it becomes 1/u(4-u)
Last thing, we need to change the limits. A limit of 4 to 1 for x means that we had to solve for x=4 and x=1. Similarly, look into the equation u = √x . You put the values of x and you get the the limits of u. E.g. u = √4 ; u = 2.
Hence, expressed in the form as given in the question.
Part b is quite easy. All you need to do is to express in the partial fractions, integrate and put in the limits. :)
Click to expand...
Need help in the one ques I posted before and
in Q.6 part (ii) of this ppr..: http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
here is the ms.: http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_ms_3.pdf
:)
 
Rutzaba

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Rutzaba

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