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Why can we differentiate trigonometric powers, but they must be simplified in order t Watch

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    When we differentiate a power of a trig function, for example sin^3x, we just imagine it as (sinx)^3 and just use the chain rule to get the answer as 3*sin^2x*cosx


    However, when integrating, I have been taught to simplify any power of a trigonometric function down until the power is just 1.

    For example, with the integral of sin^2x dx

    Simplify to: integral of 1/2 - 1/2cos2x dx

    = x/2 - (sin2x)/4

    So my question is, why must teh powers be reduced for integration, but not differentiation?

    i.e. why can sin^2x not be integrated to (sin^3x)/(6sinxcosx)?
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    the end of the title should be "to integrate?". sorry
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    (Original post by EdReed)
    the end of the title should be "to integrate?". sorry
    Because it's how integration works?

    I don't understand how you got the 6sinxcosx part...
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    (Original post by Anythingoo1)
    Because it's how integration works?

    I don't understand how you got the 6sinxcosx part...
    From 3sin(2x) I'm guessing which is incorrect.
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    (Original post by EdReed)
    When we differentiate a power of a trig function, for example sin^3x, we just imagine it as (sinx)^3 and just use the chain rule to get the answer as 3*sin^2x*cosx


    However, when integrating, I have been taught to simplify any power of a trigonometric function down until the power is just 1.

    For example, with the integral of sin^2x dx

    Simplify to: integral of 1/2 - 1/2cos2x dx

    = x/2 - (sin2x)/4

    So my question is, why must teh powers be reduced for integration, but not differentiation?

    i.e. why can sin^2x not be integrated to (sin^3x)/(6sinxcosx)?
    Well, you'd need to arrange the integral in the form nu^{n-1}\dfrac{du}{dx} for the chain rule to work backwards... you can't always do that.
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    You seem to be trying to treat sin(x) like a polynomial term. This is wrong. What you could try is this:

    let u = sin(x) \rightarrow \frac{du}{dx} = cos(x)

    \int sin^2(x) dx = \int \frac{u^2}{cos(x)} du =  \int \frac{u^2}{\sqrt{1-u^2}} du.

    EDIT: But then again to solve that integral you substitute the same thing back in so you essentially end up with your original integral anyway.
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    (Original post by EdReed)
    When we differentiate a power of a trig function, for example sin^3x, we just imagine it as (sinx)^3 and just use the chain rule to get the answer as 3*sin^2x*cosx


    However, when integrating, I have been taught to simplify any power of a trigonometric function down until the power is just 1.

    For example, with the integral of sin^2x dx

    Simplify to: integral of 1/2 - 1/2cos2x dx

    = x/2 - (sin2x)/4

    So my question is, why must teh powers be reduced for integration, but not differentiation?

    i.e. why can sin^2x not be integrated to (sin^3x)/(6sinxcosx)?
    In simple terms (because proving this is beyond Alevel, in fact beyond undergraduate degree level!), the chain rule gives you a rule that allows you to differentiate any composition of functions that you will meet at A Level: logs, trig, hyperbolics, polynomials etc

    Howver, when it comes to integration, there is no such rule - in fact, if you make up a simple looking function from a few basic functions e.g. ln(sin x) or e^{-x^2} then it's more than likely that it isn't possible to integrate it and get a result that is a finite combination of these functions!

    The "rule" that you've been taught for integration of trig powers is basically a "technique" that helps you reduce something involving sin^m x to an expression involving sin(kx) using standard trig identities, because the latter form is easy to integrate directly.

    To answer your specific example, try differentiating the thing you suggested and see if you get back to sin^2 x.
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    (Original post by davros)
    In simple terms (because proving this is beyond Alevel, in fact beyond undergraduate degree level!), the chain rule gives you a rule that allows you to differentiate any composition of functions that you will meet at A Level: logs, trig, hyperbolics, polynomials etc

    Howver, when it comes to integration, there is no such rule - in fact, if you make up a simple looking function from a few basic functions e.g. ln(sin x) or e^{-x^2} then it's more than likely that it isn't possible to integrate it and get a result that is a finite combination of these functions!

    The "rule" that you've been taught for integration of trig powers is basically a "technique" that helps you reduce something involving sin^m x to an expression involving sin(kx) using standard trig identities, because the latter form is easy to integrate directly.

    To answer your specific example, try differentiating the thing you suggested and see if you get back to sin^2 x.
    Thank you, an excellent answer.

    So I guess that, for the moment, we just have to accept it as given.

    I differentiated my example using wolfram alpha and got:

    so yes, clearly the technique does not work for integration.
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    (Original post by EdReed)
    why can sin^2x not be integrated to (sin^3x)/(6sinxcosx)?
    Try differentiating (sin^3x)/(6sinxcosx) and see what you get?
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    (Original post by Mr M)
    Try differentiating (sin^3x)/(6sinxcosx) and see what you get?
    Yes, I did that just above your post and it shows that the method does not work. However, it does not explain why.
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    try integrating sin^2 (x) then.
    sometimes in maths, you have to use alternative methods to get your answers an thats okay.
    I am sure there is a proper proof of that though.
    thing is i have never seen an actual graph for trig functions of orders more than one, so how are you gonna find the area under a non-existing graph...
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    (Original post by thephysicsguy)
    try integrating sin^2 (x) then.
    sometimes in maths, you have to use alternative methods to get your answers an thats okay.
    I am sure there is a proper proof of that though.
    thing is i have never seen an actual graph for trig functions of orders more than one, so how are you gonna find the area under a non-existing graph...
    what makes you think it's "non-existing"
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    (Original post by EdReed)
    Yes, I did that just above your post and it shows that the method does not work. However, it does not explain why.
    What you are proposing is not the same as the reverse chain rule.
    To do so you need the derivative of the function to be next to the original function.
    e.g.
    \int 2xe^{x^2} \ dx = e^{x^2} + c

    (Original post by thephysicsguy)
    try integrating sin^2 (x) then.
    sometimes in maths, you have to use alternative methods to get your answers an thats okay.
    I am sure there is a proper proof of that though.
    thing is i have never seen an actual graph for trig functions of orders more than one, so how are you gonna find the area under a non-existing graph...
    https://www.google.co.uk/search?q=si...Da720gXg3IG4Bg
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    It's quite simple. Chain rule differentiates the nested function f(g(x)) into the product of functions f'(g(x))*g'(x). Thus, "reverse chain rule" can "legally" only integrate the product of functions f'(g(x))*g'(x) into the nested function f(g(x)).

    So, is sin^2(x) in the form f'(g(x))*g'(x)? No? Then applying a "reverse chain rule" won't help you integrate it.

    Do you know any differentiation rules that always differentiates something into just a nested function? No? Then for you there's no blanket method for integrating nested functions.
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    (Original post by thephysicsguy)
    try integrating sin^2 (x) then.
    sometimes in maths, you have to use alternative methods to get your answers an thats okay.
    I am sure there is a proper proof of that though.
    thing is i have never seen an actual graph for trig functions of orders more than one, so how are you gonna find the area under a non-existing graph...
    Can't you even imagine what it might look like?!
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    okay it does exist, that was my bad mr m you seem to have a deep knowledge from your high reputation. do you have a proper and fundamental disproof for this.
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    (Original post by thephysicsguy)
    okay it does exist, that was my bad mr m you seem to have a deep knowledge from your high reputation. do you have a proper and fundamental disproof for this.
    Hm? You can integrate sin^2(x), btw. Of course it's not equal to sin^3(x)*6sinxcosx, though. The proof would then be that the two functions don't agree.
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    (Original post by thephysicsguy)
    okay it does exist, that was my bad mr m you seem to have a deep knowledge from your high reputation. do you have a proper and fundamental disproof for this.
    "Disproof" of what?
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    yes you can integrate sin^10382378283193 (x) by using de moivre theorem although it might take a bit of your time.
    i am asking for something a bit more fundamental mabe with limits or something
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    (Original post by Mr M)
    "Disproof" of what?
    trig identities with powers can't being integrated directly
 
 
 
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