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    So the question is
    Intergrate Cos to the power of 3 x Dx Using the sub U = Sin X, kinda confusing this is the first type of these questions where the sub that I use isn't in the actual question, can anyone help me to get started on this?
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    Find dx in terms of x and du and you should see that there is some cancellation. Then write the remaining terms in terms of u and integrate with respect to u - which at that point is just normal integration.
    If it is a definite integral you will have to change the limits, if it is an indefinite integral then you will have to change he variable in your answer to the original variable that you started with (you could also do this with indefinite integrals-but not necessary).

    For this example u=six so
     du/dx = \cos x \Rightarrow dx= du/ \cos x .
    Now substitute this in for the dx. You should see that the cosx cancel leaving  \cos^2 x . You rewrite this in terms of u and then integrate normally.
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    (Original post by SunDun111)
    So the question is
    Intergrate Cos to the power of 3 x Dx Using the sub U = Sin X, kinda confusing this is the first type of these questions where the sub that I use isn't in the actual question, can anyone help me to get started on this?
    Don't worry about it: u = \sin x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \cos x \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} = \frac{1}{\cos x}

    So you integral is \displaystyle \int \cos^3 x \, \mathrm{d}x = \int \cos^3 x \, \frac{ \mathrm{d}x }{ \mathrm{d}u }\mathrm{d}u since \frac{\mathrm{d}x}{\mathrm{d}u} \mathrm{d}u = \mathrm{d}x, so it's the exact same thing.

    Sub in: \displaystyle \int \cos^3 x \times \frac{1}{\cos x}  \, \mathrm{d}x = \int \cos^2 x \, \mathrm{d}u = \int 1-\sin^2 x\, \mathrm{d}x = \int 1-u^2 \, \mathrm{d}u
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    (Original post by Zacken)
    Don't worry about it: u = \sin x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \cos x \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} = \frac{1}{\cos x}

    So you integral is \displaystyle \int \cos^3 x \, \mathrm{d}x = \int \cos^3 x \, \frac{ \mathrm{d}x }{ \mathrm{d}u }\mathrm{d}u since \frac{\mathrm{d}x}{\mathrm{d}u} \mathrm{d}u = \mathrm{d}x, so it's the exact same thing.

    Sub in: \displaystyle \int \cos^3 x \times \frac{1}{\cos x}  \, \mathrm{d}x = \int \cos^2 x \, \mathrm{d}u = \int 1-\sin^2 x\, \mathrm{d}x = \int 1-u^2 \, \mathrm{d}u
    (Original post by B_9710)
    Find dx in terms of x and du and you should see that there is some cancellation. Then write the remaining terms in terms of u and integrate with respect to u - which at that point is just normal integration.
    If it is a definite integral you will have to change the limits, if it is an indefinite integral then you will have to change he variable in your answer to the original variable that you started with (you could also do this with indefinite integrals-but not necessary).

    For this example u=six so
     du/dx = \cos x \Rightarrow dx= du/ \cos x .
    Now substitute this in for the dx. You should see that the cosx cancel leaving  \cos^2 x . You rewrite this in terms of u and then integrate normally.

    Cheers guys i did the question, But i keep getting stuck on this one question it is
    2/ e to the power of 2x +4
    the U = e to the power of 2x + 4
    I can't seem to do it because I cant get rid of the DU/2e2x.
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    (Original post by SunDun111)
    Cheers guys i did the question, But i keep getting stuck on this one question it is
    2/ e to the power of 2x +4
    the U = e to the power of 2x + 4
    I can't seem to do it because I cant get rid of the DU/2e2x.
    So  \displaystyle \frac{du}{dx}=2e^{2x+4} \Rightarrow dx=\frac{du}{2e^{2x+4}}
    It doesn't cancel but now try and rewrite the term in x in the integrand in terms of u, using the substitution you have been given.
    You will have  \displaystyle \int \frac{1}{e^{4x+8}} dx =\int\frac{1}{(e^{2x+4})^2} dx .
    Just write the integrand in terms of u and integrate as you would normally.
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    (Original post by B_9710)
    So  \displaystyle \frac{du}{dx}=2e^{2x+4} \Rightarrow dx=\frac{du}{2e^{2x+4}}
    It doesn't cancel but now try and rewrite the term in x in the integrand in terms of u, using the substitution you have been given.
    You will have  \displaystyle \int \frac{1}{e^{4x+8}} dx =\int\frac{1}{(e^{2x+4})^2} dx .
    Just write the integrand in terms of u and integrate as you would normally.
    The +4 is not with the 2x its a seperate number
    So its literally 2/2e to the power of 2x + 4 as a seperate constant not with the power.
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    (Original post by SunDun111)
    The +4 is not with the 2x its a seperate number
    So its literally 2/2e to the power of 2x + 4 as a seperate constant not with the power.
     \displaystyle \int \frac{2}{e^{2x}+4} dx ?
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    (Original post by B_9710)
     \displaystyle \int \frac{2}{e^{2x}+4} dx ?
    Yes for some reason I struggle with turning the terms into U's for this question.
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    (Original post by SunDun111)
    Yes for some reason I struggle with turning the terms into U's for this question.
    Note that 2e^{2x}=2(e^{2x}+4)-8.
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    (Original post by B_9710)
     \displaystyle \int \frac{2}{e^{2x}+4} dx ?
    Ive literally done every single quetion on integreation by substition except this one, I just cant seem to do it, when I look at the mark scheme they is a 0.5x in it and it confuses me more.
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    (Original post by SunDun111)
    Ive literally done every single quetion on integreation by substition except this one, I just cant seem to do it, when I look at the mark scheme they is a 0.5x in it and it confuses me more.
    Use  u = 2e^{2x} + 4 then do partial fractions.
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    Could use  u=e^{2x} .
 
 
 
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