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  1. #1
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    integrate

    integral x^3[sqrt(9+x^2)]

    i know i am supposed to use trig sub but i am having a very confusing time trying to remember how its done.....
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  2. #2
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    Quote Originally Posted by jeph View Post
    integral x^3[sqrt(9+x^2)]

    i know i am supposed to use trig sub but i am having a very confusing time trying to remember how its done.....
    \int x^3 \sqrt{9+x^2} dx = \int \frac{x^3(9+x^2)^{3/2}}{1}\cdot \frac{1}{9+x^2}dx
    Let,
    t=\tan^{-1} (x/3) then t' = \frac{1}{9+x^2}
    And, \sec t = \frac{\sqrt{9+x^2}}{3} and \sin t = \frac{x}{3}
    Thus, 27\sec^3t=(9+x^2)^{3/2} and 27\sin^3t = x^3
    After this substitution we have,
    \int 27^2 \sec^3 t \sin^3 t dt = 729 \int \tan^3 tdt
    Can you do it frum heir?
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  3. #3
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    Let's set u=9+x^2\implies du=2x\,dx, this yields


    \begin{aligned}<br />
\int {x^3 \sqrt {9 + x^2 }\,dx} &= \frac{1}<br />
{2}\int {(u - 9)u^{1/2}\,du}\\<br />
&= \frac{1}<br />
{2}\left( {\int {u^{3/2}\,du} - 9\int {u^{1/2}\,du} } \right)\\<br />
&= \frac{1}<br />
{2}\left( {\frac{2}<br />
{5}u^{5/2} - 9 \cdot \frac{2}<br />
{3}u^{3/2} } \right) + c\\<br />
&= \frac{1}<br />
{5}(9 + x^2 )^{5/2} - 3(9 + x^2 )^{3/2} + c<br />
\end{aligned}

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