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Math Help - trig integration

  1. #1
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    trig integration

    Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

    Can someone please show where I am going wrong?

    Is my u substitution wrong?

    Thank you
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  2. #2
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    Quote Originally Posted by slaypullingcat View Post
    Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

    Can someone please show where I am going wrong?

    Is my u substitution wrong?

    Thank you
    you are not "wrong".

    try graphing your solution and the book's solution ... what do you notice?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by slaypullingcat View Post
    Hi, I can't integrate sin x cos x. I tried a u substitution with u=sinx and dx=cos x du and got 1/2(sin x)^2+c. My text book answer is -1/4cos 2x+c.

    Can someone please show where I am going wrong?

    Is my u substitution wrong?

    Thank you

    u=sinx

     <br />
du=cosx\cdot{dx}\Rightarrow{dx}=\frac{du}{cosx}<br />
Right? I don't know, I'm asking.

    Nevermind, looks like you edited your post.
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  4. #4
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    Quote Originally Posted by skeeter View Post
    you are not "wrong".

    try graphing your solution and the book's solution ... what do you notice?
    Thank you for that, but I still don't know how to integrate the 'text book' way. Can you please even suggest the method they have used or a different approach I should try?
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  5. #5
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    \int \sin{x} \cos{x} \, dx

    \frac{1}{2} \int 2\sin{x} \cos{x} \, dx

    \frac{1}{2} \int \sin(2x) \, dx


    can you get the textbook solution from this?
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  6. #6
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    Quote Originally Posted by skeeter View Post
    \int \sin{x} \cos{x} \, dx

    \frac{1}{2} \int 2\sin{x} \cos{x} \, dx

    \frac{1}{2} \int \sin(2x) \, dx


    can you get the textbook solution from this?
    Yes I can find it now. I am not very good at recognising trig identities. Thank you for your post, it makes sense now. Cheers.
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