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Thread: intergration by substitution

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    intergration by substitution

    Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

    substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]
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    Quote Originally Posted by Geordie-boy View Post
    Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

    substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]
    separate the integral into two integrals.

    $\displaystyle \int \frac{x}{\sqrt{1-x^2}} + \int\frac{x^2}{\sqrt{1-x^2}}$

    for the first integral just use the sub of $\displaystyle u= 1 - x^2$ for the second use trig sub
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  3. #3
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    Quote Originally Posted by Geordie-boy View Post
    Hi all im trying to get my head around intergration by substitution and this one problem in particular has been giving me a problems any help?

    substitute x = sine z into [int(x+x^2)/(1-x^2)^1/2 dx]
    Easy.

    You're trying to solve $\displaystyle \int{\frac{x + x^2}{\sqrt{1 - x^2}}\,dx}$ by substituting $\displaystyle x = \sin{z}$.

    Notice that if you try to do this, you need to change the integral so it is with respect to $\displaystyle z$ instead of $\displaystyle x$.

    To do this, if $\displaystyle x = \sin{z}$ then $\displaystyle \frac{dx}{dz} = \cos{z}$.

    Thus $\displaystyle dx = \cos{z}\,dz$.


    Now substitute this all into the original integral, so that it becomes

    $\displaystyle \int{\frac{\sin{z} + \sin^2{z}}{\sqrt{1 - \sin^2{z}}}\,\cos{z}\,dz}$

    $\displaystyle = \int{\frac{\sin{z} + \sin^2{z}}{\sqrt{\cos^2{z}}}\,\cos{z}\,dz}$

    $\displaystyle = \int{\frac{\sin{z} + \sin^2{z}}{\cos{z}}\,\cos{z}\,dz}$

    $\displaystyle = \int{\sin{z} + \sin^2{z}\,dz}$

    $\displaystyle = \int{\sin{z} + \frac{1}{2} - \frac{1}{2}\cos{(2z)}\,dz}$

    $\displaystyle = -\cos{z} + \frac{1}{2}z - \frac{1}{4}\sin{(2z)} + C$

    $\displaystyle = \frac{1}{2}z - \frac{1}{4}\sin{(2z)} - \cos{z} + C$.


    Now use the fact that $\displaystyle x = \sin{z}$ to get the answer back into terms of $\displaystyle x$.


    If $\displaystyle x = \sin{z}$ then $\displaystyle z = \arcsin{x}$.


    Also $\displaystyle x^2 = \sin^2{z}$

    $\displaystyle x^2 = 1 - \cos^2{z}$

    $\displaystyle x^2 - 1 = -\cos^2{z}$

    $\displaystyle 1 - x^2 = \cos^2{z}$

    $\displaystyle \sqrt{1 - x^2} = \cos{z}$.


    Finally, $\displaystyle \sin{(2z)} = 2\sin{z}\cos{z}$

    $\displaystyle = 2x\sqrt{1 - x^2}$.


    So this means that your answer is

    $\displaystyle \frac{1}{2}\arcsin{z} - \frac{1}{2}x\sqrt{1 - x^2} - \sqrt{1 - x^2} + C$.
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