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simple... ((2^6)) + 6c1 (2^4) (-x^1) bus thats as much as we are going to need
Thank u so very much....It's awesome...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...?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.
In Q.7. part 1, you need to pull up through 'integration by substitution method'.Can u plz also help me with Q.7 of the same ppr...?
Thank u..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 andIn 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.
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