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    Hi guys, really struggling with these simple questions although I can do the 'harder' questions.



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    These questions don't require any real tricks, you just need to know your derivatives of trig functions. Some of them will probably be in your formula book in fact. If you know the derivatives of tan x, sec x, cosec x and cot x then you'll be able to do these questions. The only snag is the fact that you have 5x or 7x instead of just x, but you should know how to get past that.
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    What is a list of the integrals I need to revise? I mean if the equation wants the integral of sec5xtan5x and I know the integral of sec5x and the integral of tan5x how do I apply this to the original question? Thanks a lot mate
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    (Original post by Mousebudden)
    What is a list of the integrals I need to revise? I mean if the equation wants the integral of sec5xtan5x and I know the integral of sec5x and the integral of tan5x how do I apply this to the original question? Thanks a lot mate
    These are from the Edexcel Formula Booklet
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    First one put them in terms of sin and cos than use substitution of u= cosx this will cancel out sin x and than the answer should be obvious. Same with the second one except it's now u = sinx. For last one its the cos2x double angle formula
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    (Original post by nuodai)
    These questions don't require any real tricks, you just need to know your derivatives of trig functions. Some of them will probably be in your formula book in fact. If you know the derivatives of tan x, sec x, cosec x and cot x then you'll be able to do these questions. The only snag is the fact that you have 5x or 7x instead of just x, but you should know how to get past that.
    Are you sure? because I am pretty sure what you are saying is wrong
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    Just realised you could also use trig identities
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    Integral of sec(5x)tan(5x)
    1. Set u =5x
    2. Thus du = 5dx
    3. Substitute:
    Integral of (1/5)sec(u)tan(u)du
    4. Set v = sec(u)
    5. Thus dv = sec(u)tan(u)
    6. Integral of (1/5)(1dv)
    7. Integral of (1/5)(1dv) = (1/5)v
    8. Substitute v back in.
    =(1/5)sec(u) + C
    9. Substitute u back in.
    =(1/5)sec(5x) + C
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    Integral of csc(7x)cot(7x)
    1. Set u = 5x
    2. Thus du = 5dx
    3. Substitute:
    Integral of (1/7)csc(u)cot(u)
    4. Set v = csc(u)
    5. Thus dv = -csc(u)cot(u)
    6. Integral of (-1/7)(1dv)
    7. Integral of (-1/7)(1dv) = (-1/7)v
    8. Substitute v back in.
    =(-1/7)csc(u) + C
    9. Substitute u back in.
    = (-1/7)csc(7x) + C
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    Just realised they are actually in the formula booklet. Oh well it works as proof
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    (Original post by loki276)
    Are you sure? because I am pretty sure what you are saying is wrong
    Yes I'm sure. The derivative of \sec 5x is 5\sec 5x \tan 5x, for example, and the other two integrands are also constant multiples of derivatives of simple trig functions.
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    (Original post by nuodai)
    Yes I'm sure. The derivative of \sec 5x is 5\sec 5x \tan 5x, for example, and the other two integrands are also constant multiples of derivatives of simple trig functions.
    Yeah just saw it
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    Integral of (csc(7x))^(2)
    1. Set u = 7x
    2. Thus du = 7dx
    3. Substitute:
    Integral of ((1/7)csc(u))^(2)
    **Integral of (csc(u))^(2) = -cot(u)**
    = (1/7)-cot(u)
    4. Substitute u back in.
    = -(1/7)cot(7x)
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    What I don't understand is how will the derivatives help me to find the integral? Sorry if it sounds really stupid.


    (Original post by CaliforniaKid)
    Integral of csc(7x)cot(7x)
    1. Set u = 5x
    2. Thus du = 5dx
    3. Substitute:
    Integral of (1/7)csc(u)cot(u)
    4. Set v = csc(u)
    5. Thus dv = -csc(u)cot(u)
    6. Integral of (-1/7)(1dv)
    7. Integral of (-1/7)(1dv) = (-1/7)v
    8. Substitute v back in.
    =(-1/7)csc(u) + C
    9. Substitute u back in.
    = (-1/7)csc(7x) + C
    I'm with you up untill Step 3, when you take out 1/7. How are you able to do that?

    Thanks a lot guys! All Repped
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    Keep in mind the method I used in the first two problem is the long way.

    By now you should know:
    the integral of secxtanx = secx
    the integral of cscxcotx = -cscx

    (Original post by Mousebudden)
    What I don't understand is how will the derivatives help me to find the integral? Sorry if it sounds really stupid.

    I'm with you up untill Step 3, when you take out 1/7. How are you able to do that?

    Thanks a lot guys! All Repped
    The integral is the opposite of the derivative.
    Example:
    Derivative of x^3
    =3x^2
    Integral of 3x^2
    =x^3

    I multiplied by (1/7) because there is a (7x). If there was a 7 in front of the problem I would not need to multiply by (1/7).
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    Its just the reverse of it if that makes sense
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    The integrals of those are in your formula book. I mean obviously its useful to remember them but, they are the easiest integrations you can get asked.
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    As the derivative of sec(x) is sec(x)tan(x), the integral of sec(x)tan(x) is sec(x)?

    Therefore, to refer to the original question, the integral of sec(5x)tan(5x) is sec(5x) ?
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    (Original post by Mousebudden)
    As the derivative of sec(x) is sec(x)tan(x), the integral of sec(x)tan(x) is sec(x)?

    Therefore, to refer to the original question, the integral of sec(5x)tan(5x) is sec(5x) ?
    Not quite, you're out by a factor. Note that \dfrac{d}{dx}[\sec (5x)] = 5\sec (5x)\tan (5x).
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    (Original post by Farhan.Hanif93)
    Not quite, you're out by a factor. Note that \dfrac{d}{dx}[\sec (5x)] = 5\sec (5x)\tan (5x).
    Got it. Thanks mate.
 
 
 
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