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C4 Trig. Integration Question Watch

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    Name:  image.jpg
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Size:  523.1 KB This question (Q64) is from the review exercise in the Edexcel textbook.

    I managed to do part (a) fine and got (xsin2x)/2 + (cos2x)/4 + c as the answer.

    How do you do part b?

    Thanks in advance
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    (Original post by jasminetwine)
    Name:  image.jpg
Views: 136
Size:  523.1 KB This question (Q64) is from the review exercise in the Edexcel textbook.

    I managed to do part (a) fine and got (xsin2x)/2 + (cos2x)/4 + c as the answer.

    How do you do part b?

    Thanks in advance
    Use the hint they give you !
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    (Original post by jasminetwine)
    Name:  image.jpg
Views: 136
Size:  523.1 KB This question (Q64) is from the review exercise in the Edexcel textbook.

    I managed to do part (a) fine and got (xsin2x)/2 + (cos2x)/4 + c as the answer.

    How do you do part b?

    Thanks in advance
    You can write you integral as \displaystyle \int x \left(\frac{\cos 2x + 1}{2}\right) \, \mathrm{d}x by re-arranging the given identity. This should now be easy to integrate using the answer for your first part.
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    (Original post by Zacken)
    You can write you integral as \displaystyle \int x \left(\frac{\cos 2x + 1}{2}\right) \, \mathrm{d}x by re-arranging the given identity. This should now be easy to integrate using the answer for your first part.
    Second that. Dismantle \displaystyle \int x \left(\frac{\cos 2x + 1}{2}\right) \, \mathrm{d}x to give \displaystyle     \frac{1}{2}    \int x \left(cos2x\right) \, \mathrm{d}x +          \int  \left\frac{x}{2}\right\, \mathrm{d}x                   ; by inspection the first integral is pretty much about retrieving your answer from part (a) (do remember to multiply by 1/2 though), while the second integral is something super fundamental.

    Hope this clarifies. Peace.
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    Thanks for your help so far guys!

    I can see that cos^2(x) = (cos2x+1)/2, using the identity they gave!

    Where do you go from there (i.e. How do you 'dismantle' it)?

    Thanks
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    Darling I just demonstrated this in my earlier post. Break up the original integral into two parts. Peace.
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    (Original post by WhiteGroupMaths)
    Second that. Dismantle \displaystyle \int x \left(\frac{\cos 2x + 1}{2}\right) \, \mathrm{d}x to give \displaystyle     \frac{1}{2}    \int x \left(cos2x\right) \, \mathrm{d}x +          \int  \left\frac{x}{2}\right\, \mathrm{d}x                   ; by inspection the first integral is pretty much about retrieving your answer from part (a) (do remember to multiply by 1/2 though), while the second integral is something super fundamental.

    Hope this clarifies. Peace.
    Name:  image.jpg
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Size:  275.3 KB Is this correct?
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    Perfectly done.

    Ouch my neck.

    Peace.
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    (Original post by WhiteGroupMaths)
    Perfectly done.

    Ouch my neck.

    Peace.
    Thank you so so much!

    Sorry about that!

    Have a lovely weekend
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    You too darling. God bless.
 
 
 
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