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    Hi guys
    I'm trying to integrate sin^2x dx using cos double angle formula but i keep getting wrong answers that what the text book says can you please see my working and help me if I'm wrong anywhere in it
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    (Original post by Alen.m)
    Hi guys
    I'm trying to integrate sin^2x dx using cos double angle formula but i keep getting wrong answers that what the text book says can you please see my working and help me if I'm wrong anywhere in it
    cos(2x) divided by -2 is not -cos(x), it is (-1/2)cos(2x), you cannot cancel the 2 in the 2x by dividing by 2.
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    (Original post by PLM98)
    cos(2x) divided by -2 is not -cos(x), it is (-1/2)cos(2x), you cannot cancel the 2 in the 2x by dividing by 2.
    Does that mean you cant write cos(2x)=2cos(x) and 2cos(x)/ -2 will give you -cos(x) ?
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    am I too late?
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    (Original post by Alen.m)
    Does that mean you cant write cos(2x)=2cos(x) and 2cos(x)/ -2 will give you -cos(x) ?
    No
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    (Original post by Alen.m)
    Does that mean you cant write cos(2x)=2cos(x) and 2cos(x)/ -2 will give you -cos(x) ?
    remember doing transformations of graphs in Core 1 or Core 2 (i cant remember which)

    If you have  \displaystyle  f(x) then  \displaystyle f(ax) is a horizontal scalar eg  \displaystyle f(2x) is  \displaystyle  f(x) "squeezed" into half its original width?

    and  \displaystyle af(x) is a vertical scalar eg  \displaystyle 2f(x) "stretches" the graph into 2 times its original height?

    so you can clearly say that  \displaystyle  f(2x) \neq 2f(x) (almost all the time)

    so therefore  \displaystyle cos(2x) \neq 2cos(x)
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    (Original post by DylanJ42)
    remember doing transformations of graphs in Core 1 or Core 2 (i cant remember which)

    If you have  \displaystyle  f(x) then  \displaystyle f(ax) is a horizontal scalar eg  \displaystyle f(2x) is  \displaystyle  f(x) "squeezed" into half its original width?

    and  \displaystyle af(x) is a vertical scalar eg  \displaystyle 2f(x) "stretches" the graph into 2 times its original height?

    so you can clearly say that  \displaystyle  f(2x) \neq 2f(x) (almost all the time)

    so therefore  \displaystyle cos(2x) \neq 2cos(x)
    Got it now, thanks for the explanation
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    don't forget to use Radions
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    (Original post by the bear)
    don't forget to use Radions
    Dosent make any difference here does it? As question itself dosent have any radians involves
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    (Original post by Alen.m)
    Dosent make any difference here does it? As question itself dosent have any radians involves
    i think he means in general, definite integrals involving trig are done in radians
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    (Original post by DylanJ42)
    i think he means in general, definite integrals involving trig are done in radians
    True
 
 
 
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