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    Hi, I'm not sure I'm very good at wording this question, as I've tried to ask it before and have not been understood. I'll stiill give it a shot, as it is frustrating that I don't know the answer; <.

    When I was taught how to differentiate, was I was told to multiply by the value of the indice, then reduce the power by 1, for example.
    \frac{d}{dx} x^2 = 2x

    Then, when I was taught to intergrate, I was told to increase the power by 1, and then divide by the new power, for example
    \displaystyle\int 2x\ dx = x^2

    Then when I was taught to differentiate trigonometric functions like sin squared x, I was told to multiply by the power, reduce the power by 1, and multiply by the differential of whats "in the bracket", for example,

    \frac{d}{dx} sin^2 x\ dx = 2 sinx cosx

    However, when I integrate sin^2 x
    I don't increase the power by 1, then divide by the new power, then divide by the differential for whats "in the bracket," for example:

    \displaystyle\int sin^2 x\ dx = sin^3x/3cosx = tanx sin^2 x /3

    Why is this? In my FP2 module I've been taught ways of integrating trigonometric functions with powers etc, but why is that way that worked for everything else suddenly incorrect?
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    You don't work out every integral the same way. For example integrating e^x just gives you e^x again, or integrating the natural log is something else to think about. I don't understand, it's like asking why is 1+1 equal to 2. Do you want us to explain why the integral of sin^2(x) is equal to x/2 - sin(2x)/4? I suppose you could think of it like "What would I need to differentiate to get sin^2(x)?"
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    Your teacher can't explain that?

    It's because "sin x" or something similar, such as cot x or cosh x is not a linear function (i.e. does not have a constant derivative). If you differentiate sin^3(x)/3cos(x) you must use the quotient rule, the 1/3cos(x) cannot be taken outside the differential operator. As you said, when you differentiate something like sin^2(x) you multiply by the derivative of what's in the bracket: so if you're differentiating (2x+1)^5, then the derivative of thing in the bracket is 2: a constant, and reversing this differentiation is no problem.
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    (Original post by jamie092)
    Hi, I'm not sure I'm very good at wording this question, as I've tried to ask it before and have not been understood. I'll stiill give it a shot, as it is frustrating that I don't know the answer; <.

    When I was taught how to differentiate, was I was told to multiply by the value of the indice, then reduce the power by 1, for example.
    \frac{d}{dx} x^2 = 2x

    Then, when I was taught to intergrate, I was told to increase the power by 1, and then divide by the new power, for example
    \displaystyle\int 2x\ dx = x^2

    Then when I was taught to differentiate trigonometric functions like sin squared x, I was told to multiply by the power, reduce the power by 1, and multiply by the differential of whats "in the bracket", for example,

    \frac{d}{dx} sin^2 x\ dx = 2 sinx cosx

    However, when I integrate sin^2 x
    I don't increase the power by 1, then divide by the new power, then divide by the differential for whats "in the bracket," for example:

    \displaystyle\int sin^2 x\ dx = sin^3x/3cosx = tanx sin^2 x /3

    Why is this? In my FP2 module I've been taught ways of integrating trigonometric functions with powers etc, but why is that way that worked for everything else suddenly incorrect?
    You've gotten to FP2 without understanding the chain rule from C3?

    I'm not too sure why you're considering the integral of \sin ^2x after considering it's derivative, the two results are not related.

    To integrate sin ^2x, you need to note that \cos 2x = 1-2\sin ^2x and rearrange for sin ^2x. Then try to integrate that.

    The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand. The more formal way of writing this is:

    \displaystyle\int f'(g(x))g'(x)dx = f(g(x))+C
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    Because the rules you have learnt are only applied to very specific cases.

    Oh and Farhan, if you actually read his post he isn't asking how... just why it doesn't work.
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    (Original post by Kasc)
    Because the rules you have learnt are only applied to very specific cases.

    Oh and Farhan, if you actually read his post he isn't asking how... just why it doesn't work.
    I was aware of that - that's why I answered his question and also explained to him how the integral should be dealt with.
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    (Original post by Farhan.Hanif93)
    You've gotten to FP2 without understanding the chain rule from C3?

    I'm not too sure why you're considering the integral of \sin ^2x after considering it's derivative, the two results are not related.

    To integrate sin ^2x, you need to note that \cos 2x = 1-2\sin ^2x and rearrange for sin ^2x. Then try to integrate that.

    The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand. The more formal way of writing this is:

    \displaystyle\int f'(g(x))g'(x)dx = f(g(x))+C
    Well I actually studied FP2 and C3 simultaneously, and I've often studied modules "the wrong way around," meaning I've often learned results before I've been able to derive them, which has confused me a bit. But as far as I know, I understand the chain rule from c3, but if I went the long way about differentiating sin^2 x I'd do it like this:

    y=sin^2 x

let u=sinx

y=u^2
    \frac{dy}{du} = 2u
    \frac{du}{dx} = cosx
    \frac{dy}{dx} = 2sinx cosx

    but what I was trying to say in my first post is that this rule of thumb thing where you differentiate (sinx)^2 by multiplying by the power then reduce the power by 1, and multiply by the differential of the bracket seems to work (I have multiplied by a function of x here), yet it doesn't work for integration, the other way around.

    I can definitley see why it does work for differentiation, it's just a shortened version of the chain rule. You say "The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand," but you're happy to do this the other way around with differentiation. Why is that?

    Also thanks for the help so far everyone =)
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    (Original post by jamie092)
    Well I actually studied FP2 and C3 simultaneously, and I've often studied modules "the wrong way around," meaning I've often learned results before I've been able to derive them, which has confused me a bit. But as far as I know, I understand the chain rule from c3, but if I went the long way about differentiating sin^2 x I'd do it like this:

    y=sin^2 x

let u=sinx

y=u^2
    \frac{dy}{du} = 2u
    \frac{du}{dx} = cosx
    \frac{dy}{dx} = 2sinx cosx

    but what I was trying to say in my first post is that this rule of thumb thing where you differentiate (sinx)^2 by multiplying by the power then reduce the power by 1, and multiply by the differential of the bracket seems to work (I have multiplied by a function of x here), yet it doesn't work for integration, the other way around.

    I can definitley see why it does work for differentiation, it's just a shortened version of the chain rule. You say "The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand," but you're happy to do this the other way around with differentiation. Why is that?

    Also thanks for the help so far everyone =)
    I see your problem.

    sinx is a power series. sinx=x-x^3/3! + x^5 / 5! e.t.c.

    sinx^2=(x-x^3/3! + x^5 / 5-....)^2

    That is different from say standard x^2 or x^(2/3).

    In general differentiation is easier. e^(x^2) for example is differentiable but not integrable over standard functions.
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    (Original post by jamie092)
    Hi, I'm not sure I'm very good at wording this question, as I've tried to ask it before and have not been understood. I'll stiill give it a shot, as it is frustrating that I don't know the answer; <.

    When I was taught how to differentiate, was I was told to multiply by the value of the indice, then reduce the power by 1, for example.
    \frac{d}{dx} x^2 = 2x

    Then, when I was taught to intergrate, I was told to increase the power by 1, and then divide by the new power, for example
    \displaystyle\int 2x\ dx = x^2

    Then when I was taught to differentiate trigonometric functions like sin squared x, I was told to multiply by the power, reduce the power by 1, and multiply by the differential of whats "in the bracket", for example,

    \frac{d}{dx} sin^2 x\ dx = 2 sinx cosx

    However, when I integrate sin^2 x
    I don't increase the power by 1, then divide by the new power, then divide by the differential for whats "in the bracket," for example:

    \displaystyle\int sin^2 x\ dx = sin^3x/3cosx = tanx sin^2 x /3

    Why is this? In my FP2 module I've been taught ways of integrating trigonometric functions with powers etc, but why is that way that worked for everything else suddenly incorrect?
    Google the chain rule, product rule and quotient rule. Or ask your teacher to teach you, alot of these different ways of differentiating/integrating are derived from these rules.
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    (Original post by jamie092)
    Well I actually studied FP2 and C3 simultaneously, and I've often studied modules "the wrong way around," meaning I've often learned results before I've been able to derive them, which has confused me a bit. But as far as I know, I understand the chain rule from c3, but if I went the long way about differentiating sin^2 x I'd do it like this:

    y=sin^2 x

let u=sinx

y=u^2
    \frac{dy}{du} = 2u
    \frac{du}{dx} = cosx
    \frac{dy}{dx} = 2sinx cosx

    but what I was trying to say in my first post is that this rule of thumb thing where you differentiate (sinx)^2 by multiplying by the power then reduce the power by 1, and multiply by the differential of the bracket seems to work (I have multiplied by a function of x here), yet it doesn't work for integration, the other way around.

    I can definitley see why it does work for differentiation, it's just a shortened version of the chain rule. You say "The "add one to the power and divide by the new power, then divide by derivative of the bracket" rule only applies when the derivative of the inner function is a constant or some multiple of it's derivative is already a factor of the integrand," but you're happy to do this the other way around with differentiation. Why is that?

    Also thanks for the help so far everyone =)
    Integrating in the way you're talking about only applies to a very limited number of cases that I have outlined.

    I've also explained why you can't just add to the power, divide by the new power and divide by the derivative of the bracket - There's a condition that MUST hold for your rule to be used. The simple answer is that, since integration and differentiation are inverse operations, integrating \sin ^2x and differentiating the answer should bring you back to sin ^2x. If you integrate using the method you're thinking of, and then differentiate the result, you will not get \sin ^2x. Where if you integrate it using the method I outlined at the start and differentiated the result, you will get \sin ^2x.

    If that isn't enough justification for you, google integration from first principles and try to prove that \displaystyle\int sin ^2x dx is not as simple as you're trying to make it to be. (Mind you, I don't know how difficult this could be...)
 
 
 
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