• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
4,493
Reaction score
15,418
Points
523
Ok u simplify it frst by taking out the conjugate and multiplying this as well as dividinh this by it
 
Messages
112
Reaction score
34
Points
38
Can someone plz post the notes for Integration of the sin,cos and tan etc...(P3)
Thanks in advance...
 
Messages
1,882
Reaction score
1,331
Points
173

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. :)
 
Messages
112
Reaction score
34
Points
38
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. :)
Thank u so very much....It's awesome...:) (y)
 
Messages
112
Reaction score
34
Points
38
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. :)
Can u plz also help me with Q.7 of the same ppr...? :D
 
Messages
1,882
Reaction score
1,331
Points
173
Can u plz also help me with Q.7 of the same ppr...? :D
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. :)
 
Messages
112
Reaction score
34
Points
38
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. :)
Thank u..:)
 
Messages
57
Reaction score
29
Points
18
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
 
Messages
24
Reaction score
11
Points
13
can some1 help in solving 16b and 17a,c,d
 

Attachments

  • SDC12235.JPG
    SDC12235.JPG
    1.4 MB · Views: 3
  • SDC12236.JPG
    SDC12236.JPG
    1.4 MB · Views: 4
Messages
112
Reaction score
34
Points
38
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. :)
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
:)
 
Messages
4,493
Reaction score
15,418
Points
523
Messages
213
Reaction score
188
Points
53
Messages
4,493
Reaction score
15,418
Points
523
Messages
213
Reaction score
188
Points
53
Top