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Math Help - [SOLVED] Where Am I Going Wrong In Solving this Indefinite Integral

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    [SOLVED] Where Am I Going Wrong In Solving this Indefinite Integral

    I'm not allowed to use trig substitutions to solve this, but I can use any other method. It's the integral of x/sqrt[16-x^2], I was plugging in 16-x^2 as u and -2x as du, so then it becomes -1/2 the integral of 1/sqrt[u] which simplifies to be -1/2sqrt[16-x^2] but I know from using the trig substitution method that the correct answer is actually only -sqrt[16-x^2], so where am I going wrong?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fattydq View Post
    I'm not allowed to use trig substitutions to solve this, but I can use any other method. It's the integral of x/sqrt[16-x^2], I was plugging in 16-x^2 as u and -2x as du, so then it becomes -1/2 the integral of 1/sqrt[u] which simplifies to be -1/2sqrt[16-x^2] but I know from using the trig substitution method that the correct answer is actually only -sqrt[16-x^2], so where am I going wrong?
    you have to integrate! you didn't integrate! remember, the purpose of substitutions is to get an easier integral to integrate, they don't actually save you from integration, they just make it easier.

    you are correct in getting to - \frac 12 \int \frac 1{\sqrt u}~du. note that this is - \frac 12 \int u^{-1/2}~du. now what's th integral? and what's your final answer?
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    Quote Originally Posted by Jhevon View Post
    you have to integrate! you didn't integrate.

    you are correct in getting to - \frac 12 \int \frac 1{\sqrt u}~du. note that this is - \frac 12 \int u^{-1/2}~du. now what's th integral? and what's your final answer?
    HAHA just before I read this I figured out that I forgot to integrate u^-1/2 as u^1/2 OVER ONE HALF, and was just making it u^1/2. Thanks again though!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fattydq View Post
    HAHA just before I read this I figured out that I forgot to integrate u^-1/2 as u^1/2 OVER ONE HALF, and was just making it u^1/2. Thanks again though!
    well, it makes it {\color{red}-}u^{1/2}<br />
, but yeah
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